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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 mass $M$. It is suspended by a string in a liquid of density $\rho$, where it stays vertical. The upper surface of cylinder is at a depth $h$ below the liquid surface. The force on the bottom of the cylinder by the liquid is
$\rho g (V + \pi R^2)$
$Mg$
$Mg - V \rho g$
$\rho g (V + \pi R^2 h)$
Solution
$\mathrm{F}_{2}-\mathrm{F}_{1}=$ upthrust
$\therefore \mathrm{F}_{2}=\mathrm{F}_{1}+$ upthrust
$\mathrm{F}_{2}=\left(\mathrm{p}_{0}+\rho \mathrm{gh}\right) \pi \mathrm{R}^{2}+v\rho g$
$=\mathrm{p}_{0} \pi \mathrm{R}^{2}+\rho \mathrm{g}\left(\pi \mathrm{R}^{2} \mathrm{h}+\mathrm{V}\right)$
Most appropriate option is $(D).$
