Question

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

• A. $\frac{7^{1/3}}{2}$
• B. $\frac{5^{1/3}}{2}$
• C. $\frac{9^{1/3}}{2}$
• D. $\frac{3^{1/3}}{2}$

Answer 1

The correct option is A $\frac{7^{1/3}}{2}$

Let $\rho$ be the density of the material. ${\rho }_o$ be the density of water. When the sphere has just sunk in water, then

Weight of the sphere = Weight of the water displaced
$\Rightarrow \frac{4}{3}\pi (R^3-r^3){\rho }_{s}g=\frac{4}{3}\pi R^3{\rho }_{w}g$
$\Rightarrow (R^3-r^3){\rho }_s=R^3{\rho }_w\Rightarrow \frac{(R^3-r^3)}{R^3}=\frac{{\rho }_w}{{\rho }_s}$
$\Rightarrow 1-{\left(\frac{r}{R}\right)}^3=\frac{1}{8}$
$\Rightarrow \frac{r}{R}=\frac{7^{1/3}}{2}$