Figure shows the cross-sectional view of an arrangement in which a light cylinder of radius r and length $\mathcal{l}$ [perpendicular to the plane of paper] is placed in a container containing liquid of density $\rho .$Assume liquid is incompressible and non-viscous. The net force to be applied on cylinder to keep it in equilibrium is  $\frac{\rho g\mathcal{l}{r}^2}{2}\sqrt{1+\frac{{\pi }^2}{\alpha }}.$ Determine the value of  $\alpha$

The force to be exerted on cylinder to keep it in equilibrium is equal and opposite of the force applied by liquid on cylinder.

Now, we are calculating the force experienced by liquid on cylinder. For the shown element,

$$\begin{gathered} p=\rho g\times r\sin \theta \\ dF=pda=pgr\sin \theta \times [rd\theta \times \mathcal{l}] \\ dF=\rho g\mathcal{l}{r}^2\sin \theta d\theta \\ d\overrightarrow{F}=dF\cos \theta \widehat{i}+dF\sin \theta \widehat{j} \\ \overrightarrow{F}=\underset{0}{\overset{\pi /2}{\int }}dF\cos \theta \widehat{i}+\underset{0}{\overset{\pi /2}{\int }}dF\sin \theta \widehat{j} \\ =\frac{\rho g\mathcal{l}{r}^2}{2}[\widehat{i}+\frac{\pi }{2}\widehat{j}] \end{gathered}$$

So required, force $\overrightarrow{F_1}=-\overrightarrow{F}$

$\left|\overrightarrow{F_1}\right|=\frac{\rho g\mathcal{l}{r}^2}{2}\sqrt{1+\frac{{\pi }^2}{4}}$