Two identical cylindrical vessels with their | vedclass

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(D) Let $h$ be the common height when the vessels are connected. By the conservation of mass,the total volume of liquid remains constant:$A h_1 + A h_

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

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(D) Let $h$ be the common height when the vessels are connected. By the conservation of mass,the total volume of liquid remains constant:$A h_1 + A h_2 = A h + A h = 2Ah$$h = \frac{h_1 + h_2}{2}$The initial potential energy of the system is the sum of the potential energies of the liquid in each vessel. The center of gravity of the liquid in a cylindrical vessel is at half its height:$U_i = (A h_1 ho) g \frac{h_1}{2} + (A h_2 ho) g \frac{h_2}{2} = \frac{1}{2} A ho g (h_1^2 + h_2^2)$After connecting the vessels,the final height in each vessel is $h = \frac{h_1 + h_2}{2}$. The final potential energy of the system is:$U_f = 2 	imes (A h ho) g \frac{h}{2} = A ho g h^2 = A ho g \left( \frac{h_1 + h_2}{2} ight)^2 = \frac{1}{4} A ho g (h_1 + h_2)^2$The work done by gravity is the decrease in potential energy:$W = U_i - U_f = \frac{1}{2} A ho g (h_1^2 + h_2^2) - \frac{1}{4} A ho g (h_1 + h_2)^2$$W = \frac{1}{4} A ho g [2(h_1^2 + h_2^2) - (h_1^2 + h_2^2 + 2h_1 h_2)]$$W = \frac{1}{4} A ho g (h_1^2 + h_2^2 - 2h_1 h_2) = \frac{1}{4} A ho g (h_1 - h_2)^2$

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