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A hemispherical portion of a radius R is...

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

A

`Mg`

B

`Mg-Vrhog`

C

`Mg+piR^(2)hrhog`

D

`rhog(V+piR^(2)h)`

Text Solution

Verified by Experts

The correct Answer is:
D
According to Archimedes' principle upthrust weight of the fluid displaced `F_("bottom")-F_("top")=Vrhog`

`:. F_("bottom")=F_("top")+Vrhog`
`=P_(1)xxA+Vrhog`
`=(hrhog)xx(piR^(2))+Vrhog`
`=rhog[piR^(2)h+V]`
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