Q 11-09-137JEE MainJEE Main 2020 (5 Sep, Shift 1)Medium
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 $\dfrac{27}{8}$ with respect to water, the value of $r$ is:
Answer: (A) $\dfrac89R$
Just submerged: weight of shell $=$ weight of water displaced by the full outer volume.
$$\frac43\pi(R^{3} - r^{3})\cdot\frac{27}{8}\rho_w g = \frac43\pi R^{3}\rho_w g \Rightarrow R^{3} - r^{3} = \frac{8}{27}R^{3}$$
$r^{3} = \dfrac{19}{27}R^{3} \Rightarrow r = \dfrac{19^{1/3}}{3}R \approx 0.89R \approx \dfrac89R$.
Solution by Sreeraj P, M.Sc Physics