A liquid of density $\rho$ and surface tension $\sigma$ rises to a height $h$ in a capillary tube of diameter $d$. The potential energy of the liquid in the capillary tube is 

Since the mass of the liquid in the capillary tube can be considered to be concentrated at the centre of mass which is at a height h/2, the potential energy is

             $PE=mg\frac{h}{2}={\mathrm{\pi r}}^2\mathrm{h\rho g}\frac{\mathrm{h}}{2}=\frac{\mathrm{\pi }{\left(\mathrm{rhg}\right)}^2\mathrm{\rho }}{2\mathrm{g}} \left(1\right)$

Now        $\sigma =\frac{h\rho rg}{2} or rhg=\frac{2\sigma }{\rho } \left(2\right)$

Using (2) in (1), we have 

            $PE=\frac{\mathrm{\pi }}{2g}{\left(\frac{2\sigma }{\rho }\right)}^2\rho =\frac{2{\mathrm{\pi \sigma }}^2}{\rho g}$