A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is V and its mass 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 h below the liquid surface. The force on the bottom of the cylinder by the liquid is

Options

A. `Mg`
B. `Mg-vrhog`
C. `Mg+piR^(2)hrhog`
D. `rhog(V+piR^(2)h)`

Answer

D

Explanation

` B=F_(2)-F_(1)`
`F_(2)=B+F_(1)`
`F_(2)=Vrhog+(hrhogxxpiR^(2))`
`F_(2)=rhog(V+piR^(2)h)`