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 a depth $$h$$ below the liquid surface. The force on the bottom of the cylinder by the liquid is
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 a 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.
$$\rho g\left( {V + \pi {R^2}h} \right)$$
Answer :
$$\rho g\left( {V + \pi {R^2}h} \right)$$
Solution :
According to Archimedes principle
Upthrust = Wt. of fluid displaced
$$\eqalign{ & {F_{{\text{bottom}}}} - {F_{{\text{top}}}} = V\rho g \cr & \therefore {F_{{\text{bottom}}}} = {F_{{\text{top}}}} + V\rho g \cr & = {P_1} \times A + V\rho g \cr & = \left( {h\rho g} \right) \times \left( {\pi {R^2}} \right) + V\rho g \cr & = \rho g\left[ {\pi {R^2}h + V} \right] \cr} $$
According to Archimedes principle
Upthrust = Wt. of fluid displaced
$$\eqalign{ & {F_{{\text{bottom}}}} - {F_{{\text{top}}}} = V\rho g \cr & \therefore {F_{{\text{bottom}}}} = {F_{{\text{top}}}} + V\rho g \cr & = {P_1} \times A + V\rho g \cr & = \left( {h\rho g} \right) \times \left( {\pi {R^2}} \right) + V\rho g \cr & = \rho g\left[ {\pi {R^2}h + V} \right] \cr} $$
