Question

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 depth $h$ below the liquid surface. The force on the bottom of the cylinder by the liquid is:

• A. $Mg$
• B. $Mg-V\rho g$
• C. $Mg+\pi R^{2}h\rho g$
• D. $\rho g(V+\pi R^{2}h)$

Answer 1

The correct option is D $\rho g(V+\pi R^{2}h)$

As the side forces line

to pressure cancels

each other out

the buoyant force (Jgv) is equal to the differentiate

b/w force on the bottom and that

on the upper part.

Force on upper parts =pressure $\times$ Area

$Fu=\left(lgh\right)\left(\pi R^2\right)$

So,

Buoyant force =$F_B-F_a$

$F_B=Buoyantforce+Fu$

$F_B=Jgv+Jgh\pi R^2$

$=Jg(V+h\pi R^2)$

option D is correct