A cylinder of radius R, height H and density `simga` has a hemispherical cut at its bottom. The top of the cyliner is kept at depth h from the liquid surface. If the density of liquid is `rho`, find the hydrostatic force acting on the hemispherical surface of the cylinder.

Options

A. `F_(2)=piR^(2)rhog(H+h-(2)/(3)R)`
B. `F_(2)=piR^(2)rhog(H-h+(2)/(3)R)`
C. `F_(2)=piR^(2)rhog(H-h-(2)/(3)R)`
D. `F_(2)=piR^(2)rhog(H+h+(2)/(3)R)`

Answer

A

Explanation

Let us assume that the liquid presses the top and bottom of the cylinder with force `F_(1)` and `F_(2)` respectively. Since the net side thrust is zero, the net hydrostatic force, that is buoyant force is
`F_(b)=F_(2)-F_(1),F_(b)=Vrhog` where `F_(2)=Vrhog+F_(1)`
Substituting `V=` volume of the body `=piR^(2)H-(2)/(3)piR^(3)`
and `F_(1)=P_(1)A=(rhogh)(piR^(2))`
in the above equation, we have
`F_(2)=(piR^(2)H-(2)/(3)piR^(3))pirgrho+piR^(2)rhogh`
This gives `F_(2)=piR^(2)rhog(H+h-(2)/(3)R)`