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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+πR2ρhg

d

ρg(V+πR2h)

answer is D.

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Detailed Solution

F2−F1=B F2=F1+B=(Ahρg)+vρg=ρg[πR2h+v]∴  F2=ρg[v+πR2h]
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