Question
A hollow spherical shell at 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 $\frac{27}{8}$ $w.r.t.$ water, the value of $r$ is$......R$
✓
Answer
$\frac{4}{3} \pi\left( R ^{3}- r ^{3}\right) \rho_{ m } g =\frac{4}{3} \pi R ^{3} \rho_{ w } g$
$1-\left(\frac{ r }{ R }\right)^{3}=\frac{8}{27}$
$\Rightarrow \frac{ r }{ R }=\left(\frac{19}{27}\right)^{1 / 3}=\frac{19^{1 / 3}}{3}$
$=0.88\simeq \frac{8}{ g }$
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