A hemispherical bowl just floats without sink | vedclass

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(C) The weight of the hemispherical bowl is given by $W = mg = V_{material} \rho g$,where $V_{material} = \frac{2}{3} \pi [ (D/2)^3 - (d/2)^3 ]$ is th

Vedclass · Accepted Answer

$0.98$

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(C) The weight of the hemispherical bowl is given by $W = mg = V_{material} ho g$,where $V_{material} = \frac{2}{3} \pi [ (D/2)^3 - (d/2)^3 ]$ is the volume of the material of the bowl.Here,$D = 1 \ m$ is the outer diameter and $d$ is the inner diameter. The density of the bowl is $ho = 2 	imes 10^4 \ kg/m^3$.According to the principle of flotation,the weight of the bowl equals the weight of the liquid displaced by the submerged part of the bowl. Since the bowl is hemispherical and just floats,the volume of the liquid displaced is the volume of the outer hemisphere,$V_{displaced} = \frac{2}{3} \pi (D/2)^3$.The weight of the displaced liquid is $W_{liquid} = V_{displaced} \sigma g$,where $\sigma = 1.2 	imes 10^3 \ kg/m^3$ is the density of the liquid.Equating the two weights: $\frac{2}{3} \pi (D/2)^3 \sigma g = \frac{2}{3} \pi [ (D/2)^3 - (d/2)^3 ] ho g$.Canceling common terms: $(D/2)^3 \sigma = [ (D/2)^3 - (d/2)^3 ] ho$.Substituting the values: $(0.5)^3 	imes 1.2 	imes 10^3 = [ (0.5)^3 - (d/2)^3 ] 	imes 2 	imes 10^4$.$0.125 	imes 1.2 	imes 10^3 = [ 0.125 - (d/2)^3 ] 	imes 20 	imes 10^3$.$0.15 = [ 0.125 - (d/2)^3 ] 	imes 20$.$0.0075 = 0.125 - (d/2)^3$.$(d/2)^3 = 0.125 - 0.0075 = 0.1175$.$d/2 = (0.1175)^{1/3} \approx 0.49$. $d = 2 	imes 0.49 = 0.98 \ m$.

$A$ hemispherical bowl just floats without sinking in a liquid of density $1.2 \times 10^3 \ kg/m^3$. If the outer diameter and the density of the bowl are $1 \ m$ and $2 \times 10^4 \ kg/m^3$ respectively,then the inner diameter of the bowl will be........ $m$.

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