Question

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

• A. $\frac{\sqrt[3]{33}}{2}$
• B. $\frac{\sqrt[3]{35}}{2}$
• C. $\frac{\sqrt[3]{37}}{2}$
• D. $\frac{3}{\sqrt[3]{37}}$

Answer 1

The correct option is C $\frac{\sqrt[3]{37}}{2}$

Since, specific gravity of sphere is $8$,

$\therefore$ density of water $=8{\rho }_w$


As, the sphere just sinks in water

$\therefore \text{Weight of sphere = upthrust acting on sphere}$

$(\frac{4}{3}\pi R^3-\frac{4}{3}\pi r^3)(8\times {\rho }_w\times g)=\left(\frac{4}{3}\pi R^3\right){\rho }_{w}g$

$\Rightarrow 8R^3-8r^3=R^3$

$\Rightarrow 8r^3=7R^3$

$\therefore \frac{r}{R}={\left(\frac{7}{8}\right)}^{1/3}=\frac{\sqrt[3]{37}}{2}$

Hence, option $\left(\text{C}\right)$ is correct.