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  • 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 liq

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 \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 (Fig. 7.31)

                                       

Option: 1

\mathrm{M g}


Option: 2

\mathrm{M g-V \rho g}


Option: 3

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


Option: 4

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


Answers (1)

best_answer

From Archimedes' principle, \mathrm{F_b-F_t=} upthrust \mathrm{=} weight of volume V of the liquid. Here \mathrm{F_b} and \mathrm{F_t } denote the force exerted by the liquid on the bottom and the top of the cylinder respectively. Upthrust \mathrm{= \rho V g} and \mathrm{F_t=\left(\pi R^2 h\right) \rho g.} Hence

                         \mathrm{ F_b=\left(\pi R^2 h\right) \rho g+\rho V g=\rho g\left(\pi R^2 h+V\right) }
Hence the correct choice is (d).

Posted by

jitender.kumar

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