A hollow spherical shell of outer radius R fl | vedclass

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(B) For the shell to float just submerged,the weight of the shell must be equal to the weight of the water displaced.Weight of shell = $V_{material} \

Vedclass · Accepted Answer

$0.88$

Vedclass · Answer

(B) For the shell to float just submerged,the weight of the shell must be equal to the weight of the water displaced.Weight of shell = $V_{material} 	imes ho_{material} 	imes g = \frac{4}{3} \pi (R^3 - r^3) ho_{material} g$.Weight of displaced water = $V_{total} 	imes ho_{water} 	imes g = \frac{4}{3} \pi R^3 ho_{water} g$.Equating the two: $\frac{4}{3} \pi (R^3 - r^3) ho_{material} g = \frac{4}{3} \pi R^3 ho_{water} g$.$(R^3 - r^3) ho_{material} = R^3 ho_{water}$.Given specific gravity $\frac{ho_{material}}{ho_{water}} = \frac{27}{8}$.So,$(R^3 - r^3) \frac{27}{8} = R^3$.$R^3 - r^3 = \frac{8}{27} R^3$.$r^3 = R^3 - \frac{8}{27} R^3 = \frac{19}{27} R^3$.$r = R \left( \frac{19}{27} ight)^{1/3} = \frac{R}{3} (19)^{1/3}$.Since $(19)^{1/3} \approx 2.668$,$r \approx \frac{2.668}{3} R \approx 0.889 R$.

$A$ hollow spherical shell of outer radius $R$ floats just submerged under the water surface. The inner radius of the shell is $r$. If the specific gravity of the shell material is $\frac{27}{8}$ with respect to water,the value of $r$ is $......R$.

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