Two identical cylindrical vessels with their | vedclass

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(D) The work done by gravity is equal to the decrease in the potential energy of the liquid system.Initial potential energy $U_i = (m_1 g \frac{h_1}{2

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$\frac{1}{4}(h_1 - h_2)^2gAd$

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(D) The work done by gravity is equal to the decrease in the potential energy of the liquid system.Initial potential energy $U_i = (m_1 g \frac{h_1}{2}) + (m_2 g \frac{h_2}{2}) = (A h_1 d g \frac{h_1}{2}) + (A h_2 d g \frac{h_2}{2}) = \frac{Adg}{2}(h_1^2 + h_2^2)$.When connected, the final height in both vessels will be $h_{avg} = \frac{h_1 + h_2}{2}$.Final potential energy $U_f = 2 	imes (A h_{avg} d g \frac{h_{avg}}{2}) = A d g h_{avg}^2 = Adg (\frac{h_1 + h_2}{2})^2$.Work done $W = U_i - U_f = \frac{Adg}{2}(h_1^2 + h_2^2) - Adg \frac{(h_1 + h_2)^2}{4}$.$W = \frac{Adg}{4} [2(h_1^2 + h_2^2) - (h_1^2 + h_2^2 + 2h_1h_2)] = \frac{Adg}{4} (h_1^2 + h_2^2 - 2h_1h_2) = \frac{1}{4} Adg (h_1 - h_2)^2$.

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