Question:
A hollow spherical shell at outer radius $\mathrm{R}$ floats just submerged under the water surface. The inner radius of the shell is $\mathrm{r}$. If the specific gravity of the shell material is $\frac{27}{8}$ w.r.t. water, the value of $r$ is :
Correct Option: , 2
Solution:
$\frac{4}{3} \pi\left(R^{3}-r^{3}\right) \rho_{m} g=\frac{4}{3} \pi R^{3} \rho_{w} g$
$1-\left(\frac{\mathrm{r}}{\mathrm{R}}\right)^{3}=\frac{8}{27}$
$\Rightarrow \frac{\mathrm{r}}{\mathrm{R}}=\left(\frac{19}{27}\right)^{1 / 3}=\frac{19^{1 / 3}}{3}$
$=0.88 \simeq \frac{8}{9}$
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