Two identical cylindrical vessels are kept on the ground and each contain the same liquid of density $d$. The area of the base of both vessels is $S$ but the height of liquid in one vessel is $x_1$ and in the other $x_2$. When both cylinders are connected through a pipe of negligible volume very close to the bottom, the liquid flows from one vessel to the other until it comes to equilibrium at a new height. The change in energy of the system in the process is:
Answer: (D) $\dfrac14gdS(x_2 - x_1)^{2}$
A column of height $x$ has potential energy $(dSx)g\dfrac x2 = \dfrac12dSgx^{2}$.
Initially $U_i = \dfrac12dSg(x_1^{2} + x_2^{2})$. Finally both heights are $\dfrac{x_1 + x_2}{2}$: $U_f = dSg\left(\dfrac{x_1 + x_2}{2}\right)^{2}$.
$$U_i - U_f = \frac{dSg}{4}\left[2x_1^{2} + 2x_2^{2} - (x_1 + x_2)^{2}\right] = \frac14gdS(x_2 - x_1)^{2}$$
Solution by Sreeraj P, M.Sc Physics