A hemispherical portion of a radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string in a liquid of density `rho` where it stays vertical . The upper surface of the cylinder is at a depth of h below the liquid surface. THe force on the bottom of the cylinder by the liquid is

Options

A. `Mg`
B. `Mg-Vrho g`
C. `Mg +piR^(2)h rho g`
D. `rho g(V +pi R^(2)h)`

Answer

D

Explanation

`[F_("lower") -F_("upper")]by_("solid") =` Uptrust
`F_(2) -F_(1)`= upthrust
`:. F_(2) =F_(1) +` upthrust
`F_(2) = rho gh (pi R^(2))+V rho g`

or `F_(2) = rho g(V + pi R^(2) g)`
In this problem, we did not take the force due to air pressure on the cylinder. This in because force due to air pressure is cancelled. At top and bottom of the cylinder the force due to air pressure is equal and opposite.