# 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 sphecific gravity of the shell material is $\frac{27}{8}$ w.r.t water, the value of $r$ is : Option: 1 Option: 2 Option: 3 $\frac{2}{3}R$ Option: 4

$\begin{array}{l} \frac{4}{3} \pi\left(\mathrm{R}^{3}-\mathrm{r}^{3}\right) \rho_{\mathrm{m}} \mathrm{g}=\frac{4}{3} \pi \mathrm{R}^{3} \rho_{\mathrm{w}} \mathrm{g} \\ \\ 1-\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^{3}=\frac{8}{27} \\ \\ \Rightarrow \frac{\mathrm{r}}{\mathrm{R}}=\left(\frac{19}{27}\right)^{1 / 3}=\frac{19^{1 / 3}}{3} \\\\ =0.88 \simeq \frac{8}{9} \end{array}$

$\Rightarrow r = \frac{8R}{9}$

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