A hemispherical portion of radius ' $R$ ' is removed from the bottom of a cylinder of radius ' R '. The…
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 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 ' $h$ ' below the liquid surface. The force on the bottom of the liquid is
Mg
$\mathrm{Mg}-\mathrm{V} \rho \mathrm{g}$
$\mathrm{Mg}+\pi \mathrm{R}^2 \mathrm{~h} \rho g$
$\quad \rho g\left(V+\pi r^2 h\right)$
Solution
Net upward zone force on the bottom of the liquid $=$ weight of the liquid displaced by cylinder + thrust force on upper surface of the cylinder due to h column of liquid
$\begin{aligned}
& F_{\text {net }}=\rho V g+\rho g h \times \pi r^2 \\
&=\rho g\left(V+\pi r^2 h\right)
\end{aligned}$