question_answer

A hemispherical portion of radius R is removed from the bottom of a cylinder of radius R. The volume of the remaining cylinder is T and its mass M. It is suspended by a string in a liquid of density p 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  

A) \[Mg\]

B) \[Mg-V\rho g\]

C) \[Mg+\pi {{R}^{2}}h\rho g\]

D) \[\rho g(V+\pi {{R}^{2}}h)\]

Correct Answer: D

Solution :

[d] According to Archimedes principle Upthrust = Wt. of fluid displaced \[{{F}_{bottom}}-{{F}_{top}}=V\rho g\] \[\therefore \,\,\,{{F}_{bottom}}={{F}_{top}}+V\rho g\] \[={{P}_{1}}\times A+V\rho g\] \[=(h\rho g)\times (\pi {{R}^{2}})+V\rho g\] \[=\rho g[\pi {{R}^{2}}h+V]\]