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A hemispherical portion of radius R is r...

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

A

Mg

B

Mg-`vrhog`

C

`Mg+piR^2hgrho`

D

`rhog(V+piR^2h)`

Text Solution

Verified by Experts

The correct Answer is:
D
The atmospheric pressure on the top and bottom of the cylinder is equal and opposite

The force on the upper surface of the cylinder
`=F_1`=Hydrostatic pressure`xx`Area=`hrhogxxpiR^2`
The force on the lower surface`=F_2`
and `F_2-F_1`= Upthrust=Volume of the remaining cylinder`xx`density of the liquid`xx`g
`=Vxxrhoxxg`
`:." "F_2=F_1`+Upthrust=`hrhogxxpiR^2+Vrhog`
`F_2=rhog(V+piR^2h)`
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