b
Balancing force on both sides Horizontal force acting on the cylinder can be assumed to be acting on the cross$-$sectional area in the vertical direction
$2 \mathrm{ogh} . \frac{\mathrm{h}}{2}=\frac{3 \rho \mathrm{gR.} \mathrm{R}}{2}$
$h^{2}=\frac{3}{2} R^{2}$
$\mathrm{h}=\sqrt{\frac{3}{2}} \mathrm{R}$
