Q: 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 ρ . 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-Vρg

(c) Mg+ πR^{2}hρg

(d) ρg (V + πR^{2}h)

Ans: (d)

Sol: Let F_{2} = Force on bottom of cylinder

F_{1} = Force on top of cylinder

Upthrust = F_{2} – F_{1}

F_{2} = F_{1} + Upthrust

$\large F_2 = (p_0 + \rho g h )\pi R^2 + V \rho g$

$\large F_2 = p_0 \pi R^2 + \rho g ( \pi R^2 h + V ) $

Here , most suitable option is

F_{2} = ρg (V + πR^{2}h)