JEE Main · 2020 · 05 Sep 2020 · Shift 1 · Fluid Mechanics

A floating body displaces its own weight of water, so the weight of water displaced by the full outer sphere equals the weight of the shell material. Writing densities relative to water, $\frac{4}{3}\pi R^3{\rho }_w=\frac{4}{3}\pi \left(R^3-r^3\right){\rho }_s$ with ${\rho }_s=\frac{27}{8}{\rho }_w$. The common factor $\frac{4}{3}\pi {\rho }_w$ cancels, leaving $R^3=\frac{27}{8}\left(R^3-r^3\right)$, that is $8R^3=27R^3-27r^3$, so $27r^3=19R^3$ and $r^3=\frac{19}{27}{R}^3$. Taking the cube root, $r={\left(\frac{19}{27}\right)}^{1/3}R\approx 0.89R$, which is $\frac{8}{9}R$.