Question

In the figure shown, the heavy cylinder (radius $R$) resting on a smooth surface separates two liquids of densities $2\rho$ and $3\rho$. If the height '$h$' for the equilibrium of cylinder is $R\sqrt{\frac{x}{2}}$. Find $x$

Answer 1

First, lets concentrate on the force exerted by the liquid of density $3\rho$ on the cylinder in the horizontal direction.

Let the length of the cylinder be L.

Consider a small segment of length $rd\theta$ at an angle $\theta$ from the horizontal line.

Height of this segment from the topmost point of fluid $3\rho$ is $R\sin \theta$

Hence, the pressure exerted by the fluid will be $3\rho gR\sin \theta$

The force exerted in the horizontal direction, $dF=3\rho gR\sin \theta RL\cos \theta d\theta$

Hence, $F={\int }_0^{\frac{\pi }{2}}3\rho g{R}^{2}L\sin \theta \cos \theta d\theta$

$F=1.5\rho g{R}^{2}L$

Similarly, proceeding for the fluid with density $2\rho$

Height of any segment, above horizontal $=h-R-R\sin \theta$

below horizontal, $h-R+R\sin \theta$

Thus horizontal force on the cylinder because of fluid,

$F={\int }_{-\theta }^{0}2\rho g(h-R-R\sin \theta )Rd\theta L+{\int }_0^{\frac{\pi }{2}}2\rho g(h-R+R\sin \theta )Rd\theta L$

For equilibrium, both the forces should be equal, hence solving the above equation,

h = R $\sqrt{\frac{3}{2}}$

Hence, $x=3$