A water tank has a circular hole at its base. A solid cone is used to plug the hole. Exactly half the height of the cone protrudes out of the hole. Water is filled in the tank to a height equal to height of the cone. Calculate the buoyancy force on the cone. Density of water is $\mathrm{\rho }$ and volume of cone is V.

V = volume of cone $=\frac{1}{3}{\mathrm{\pi R}}^2\mathrm{h}$
Volume of protruding part ${\mathrm{V}}_{\text{out }}=\frac{1}{3}\mathrm{\pi }{\left(\frac{\mathrm{R}}{2}\right)}^2\frac{\mathrm{h}}{2}=\frac{\mathrm{V}}{8}$
Volume of cone inside water ${\mathrm{V}}_{\text{in }}=\mathrm{V}-\frac{\mathrm{V}}{8}=\frac{7\mathrm{V}}{8}$
Imagine the cone to have that part of it which protrudes through the hole removed and the space under the container filled with water. The buoyancy force would then be
${\mathrm{F}}_0={\mathrm{\rho gV}}_{\text{in }}=\frac{7\mathrm{V\rho g}}{8}$
As there is no water beneath the hole, a contribution $\mathrm{\pi }{\left(\frac{\mathrm{R}}{2}\right)}^2\mathrm{\rho gh}$ is missing
Actual buoyancy force is $\mathrm{F}={\mathrm{F}}_0-\mathrm{\pi }\frac{{\mathrm{R}}^2}{4}\mathrm{\rho gh}=\frac{7}{8}\mathrm{V\rho g}-\frac{3\mathrm{V}}{4}\mathrm{\rho g}=\frac{\mathrm{V\rho g}}{8}$