A hemispherical portion of radius a is removed from the bottom of a cylinder of radius a. The volume of the remaining cylinder is V and its mass is M. It is suspended by a string of length l in a liquid of density ρ where 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 is
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a
Mg
b
Mg−Vρg
c
Mg+πa2hρg
d
ρg(V+πa2h)
answer is D.
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Detailed Solution
Fat bottom = Fat top +up thrust =(ρgh).πa2+Vρg=ρg(πa2h+V)