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  • A hemispherical portion of radius is removed from the bottom of a cylinder of radius R. The volume of

A hemispherical portion of radius \mathrm{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 \mathrm{\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

Option: 1

\mathrm{\mathrm{Mg}}


Option: 2

\mathrm{M g-V \rho g}


Option: 3

\mathrm{\mathrm{Mg}+\pi \mathrm{R}^2 \mathrm{~h} \rho \mathrm{g}}


Option: 4

\mathrm{\rho g\left(V+\pi R^2 h\right)}


Answers (1)

best_answer

\mathrm{ \vee \rho g+\pi r^2 h \rho g }
Wt of big column & buyorny on block.

Posted by

Sanket Gandhi

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