In the Figure shown, the heavy cylinder (radius R) resting on a smooth surface separates two liquids of densities 2$\rho$ and 3$\rho$. The height h for the equilibrium of cylinder must be

Torque due to hydrostatic force about centre of sphere is zero because hydrostatic force passes through the centre. So, horizontal force on the cylinder due to two liquids must cancel out.

$\Rightarrow {\mathrm{F}}_{\mathrm{LHS}}={\mathrm{F}}_{\mathrm{RHS}}$

If L be the length of cylinder, then

$\begin{array}{l}{\left({\mathrm{P}}_{\text{avg }}\right)}_{\text{left }}\mathrm{hL}={\left({\mathrm{P}}_{\text{avg }}\right)}_{\text{right }}\mathrm{RL}\\ \Rightarrow \frac{1}{2}\left(2\mathrm{\rho gh}\right)\left(\mathrm{hL}\right)=\frac{1}{2}\left(3\mathrm{\rho gR}\right)\left(\mathrm{RL}\right)\\ \Rightarrow \mathrm{h}=\sqrt{\frac{3}{2}}\mathrm{R}\end{array}$