A sphere of solid material of specific gravity 8 has a concentric spherical cavity and just sinks in water. The ratio of radius of cavity to that of outer radius of the sphere must be

Solution :
Let $$\rho $$ be the density of the material. $${\rho _0}$$ be the density of water when the sphere has just started sinking, the weight of the sphere = weight of water displaced (approx).
$$\eqalign{ & \Rightarrow \frac{4}{3}\pi \left( {{R^3} - {r^3}} \right)\rho g = \frac{4}{3}\pi {R^3}{\rho _0}g \cr & \Rightarrow \left( {{R^3} - {r^3}} \right)\rho = {R^3}\rho 0 \Rightarrow \frac{{\left( {{R^3} - {r^3}} \right)}}{{{R^3}}} = \frac{{{\rho ^0}}}{\rho } \cr & \Rightarrow \frac{r}{R} = \frac{{{{\left( 7 \right)}^{\frac{1}{3}}}}}{2} \cr} $$