Two cylindrical vessels of equal cross-sectional area $A$ contain water upto heights $h_{1}$ and $h_{2}$ . The vessels are interconnected so that the levels in them become equal. Calculate the work done by the force of gravity during the process. The density of water is $ρ$ .

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Solution

Since total volume remains constant, the final height of the liquid is:
$\frac{h_{1} + h_{2}}{2} = H$
Initial potential energy of container
$A h_{1} ρ g \times \frac{h_{1}}{2} + A h_{2} ρ g \times \frac{h_{2}}{2} = \frac{A ρ g}{2} [h_{1}^{2} + h_{2}^{2}]$
Final potential energy
$= 2 \times A ρ (\frac{h_{1} + h_{2}}{2}) \times (\frac{h_{1} + h_{2}}{2}) g = \frac{A ρ g}{2} {(h_{1} + h_{2})}^{2}$
Work done $= P_{f} - P_{i} = \frac{A ρ g}{2} [{2 h}_{1} h_{2}]$
$W = A ρ g h_{1} h_{2}$

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Two cylindrical vessels of equal cross-sectional area A contain water upto heights h1 and h2. The vessels are interconnected so that the levels in them become equal. Calculate the work done by the force of gravity during the process. The density of water is ρ.

Since total volume remains constant, the final height of the liquid is: h1+h22=HInitial potential energy of container Ah1ρg×h12+Ah2ρg×h22=Aρg2[h21+h22]Final potential energy =2×Aρ(h1+h22)×(h1+h22)g=Aρg2(h1+h2)2Work done=Pf−Pi=Aρg2[2h1h2] W=Aρgh1h2

Two cylindrical vessels of equal cross-sectional area $A$ contain water upto heights $h_{1}$ and $h_{2}$ . The vessels are interconnected so that the levels in them become equal. Calculate the work done by the force of gravity during the process. The density of water is $ρ$ .