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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 its mass is 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

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a

Mg

b

Mg−Vρg

c

Mg+πR2hρg

d

ρg(v+πR2h)

answer is D.

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Detailed Solution

F2−F1=FB                 F2=F1+FB=(Ahρg)+vρg                =ρg(Ah+v)=ρg(πR2h+v)                =ρg(πR2h+v)                ∴  F2=ρg(v+πR2h)
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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 its mass is 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