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 ρ 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

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 ρ 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 <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/ee8dd9ba920543afa820191af3a25193-.png" /> (A) M g (B) M g-V ρ g (C) M g+π R2 h ρ g (D) ρ g(V+π R2 h)

Tardigrade Admin · Accepted Answer

F2=F1+ upthrust or F2=(ρ g h)(π R2)+V ρ g or F2=ρ g(π R2 h+V) <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/7d75e0a076271ae2f9ca89ca8f20201e-.png" />

$M g$

$M g - V ρ g$

$M g + π R^{2} h ρ g$

$ρ g (V + π R^{2} h)$

$F_{2} = F_{1} +$ upthrust
or $F_{2} = (ρ g h) (π R^{2}) + V ρ g$
or $F_{2} = ρ g (π R^{2} h + V)$

Mg

Mg−Vρg

Mg+πR2hρg

ρg(V+πR2h)

F2​=F1​+ upthrust or F2​=(ρgh)(πR2)+Vρg or F2​=ρg(πR2h+V)

$M g$

$M g - V ρ g$

$M g + π R^{2} h ρ g$

$ρ g (V + π R^{2} h)$

$F_{2} = F_{1} +$ upthrust
or $F_{2} = (ρ g h) (π R^{2}) + V ρ g$
or $F_{2} = ρ g (π R^{2} h + V)$