$\rho A_{1} h_{1}+\rho A_{2} h_{2}=\rho h\left(A_{1}+A_{2}\right)$

$h=\left(h_{1}+h_{2}\right) / 2\;\;\;\left[\right.$ as $A_{1}=A_{2}=A$ given $]$

As $\left(h_{1} / 2\right)$ and $\left(h_{2} / 2\right)$ are heights of initial centre of gravity of liquid in two vessels., the initial potential energy of the system $U_{i}=\left(h_{1} A \rho\right) g \frac{h_{1}}{2}+\left(h_{2} A \rho\right) \frac{h_{2}}{2}=\rho g A \frac{\left(h_{1}^{2}+h_{2}^{2}\right)}{2} \ldots( i )$

When vessels are connected the height of centre of gravity of liquid in each vessel will be $h / 2$, i.e. $\left(\frac{\left(h_{1}+h_{2}\right)}{4}\right.$ [as $\left.h=\left(h_{1}+h_{2}\right) / 2\right]$

Final potential energy of the system

$U_{F}=\left[\frac{\left(h_{1}+h_{2}\right)}{2} A \rho\right] g\left(\frac{h_{1}+h_{2}}{4}\right)$

$=A \rho g\left[\frac{\left(h_{1}+h_{2}\right)^{2}}{4}\right] \ldots$.(ii)

Work done by gravity $W=U_{i}-U_{f}=\frac{1}{4} \rho g A\left[2\left(h_{1}^{2}+h_{2}^{2}\right)-\left(h_{1}+h_{2}\right)^{2}\right]$

$=\frac{1}{4} \rho g A\left(h_{1}-h_{2}\right)^{2}$