Two identical cylindrical vessels with their bases at same level each contains a liquid of density r. The height of the liquid in one vessel is h₁ and that in the other vessel is h₂ . The area of either base is A. The work done by gravity in equalizing the levels when the two vessels are connected, is

Work done by gravity is equal to (change in potential energy)
assuming h to be the final equilibrium

$W_{Cylinder}=U_i-U_f$

$W_{Cylinder1}=A\rho g\frac{h_1}{2}(h_1-h)$

$W_{Cylinder2}=A\rho g\frac{h_2}{2}(h_2-h)$

Thus the total work done by gravity,

$$\begin{gathered} W_g=W_{Cylinder1}+W_{Cylinder2} \\ \Rightarrow W_g=A\rho g\frac{h_1}{2}(h_1-h)+A\rho g\frac{h_2}{2}(h_2-h) \end{gathered}$$

By using the volume of conservation,

$h=\frac{h_1+h_2}{2}$

$W_g=A\rho g\frac{h_1}{2}(h_1-\frac{h_1+h_2}{2})+A\rho g\frac{h_2}{2}(h_2-\frac{h_1+h_2}{2})$

$$\begin{gathered} W_g=A\rho g{\left(\frac{h_1+h_2}{2}\right)}^2 \\ \Rightarrow W_g=\frac{1}{4}A\rho g{\left(h_1-h_2\right)}^2 \end{gathered}$$

Hence the correct answer is $\frac{1}{4}A\rho g{\left(h_1-h_2\right)}^2.$