A cylindrical vessel of radius r is filled with a liquid to a height h such that the force exerted by the liquid on the sides of the vessel is equal to that exerted at the bottom. The ratio $\frac{h}{r}$ is

Pressure at the bottom = $h\rho g$. Therefore, the force exerted at the bottom is $F_1=h\rho g\pi r^2$

Now, since the mass of the liquid can be regarded as being concentrated at its centre of mass which is at a height h/2, the pressure on the sides =$\frac{h}{2}\rho g$

 Therefore, force exerted on sides is 

        $$\begin{gathered} F_2=\frac{h}{2}\rho g\times 2\mathrm{\pi rh}.\mathrm{Given} {\mathrm{F}}_1={\mathrm{F}}_2 \mathrm{or} \\ \mathrm{h\rho g}\times {\mathrm{\pi r}}^2=\frac{\mathrm{h}}{2}\mathrm{\rho g}\times 2\mathrm{\pi rh} \end{gathered}$$

which gives r = h. Hence the correct choice is (a).