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 r 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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Arun · Answer

Dear Sandeep Using Archimedes principle, F bottom - h&rho;g(&pi;R 2 ) = Vrg So, Fbottom = &rho;g(V + &pi;R 2 h) (ignoring atmospheric pressure) Regards Arun (askIITians forum expert)

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