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

Answer: (A)

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).
$\Rightarrow \frac{4}{3} \pi\left(R^3-r^3\right) \rho g=\frac{4}{3} \pi R^3 \rho_0 g$
$\Rightarrow\left(R^3-r^3\right) \rho=R^3 \rho 0$
$ \Rightarrow \frac{\left(R^3-r^3\right)}{R^3}=\frac{\rho^0}{\rho}$
$ \Rightarrow \frac{r}{R}=\frac{(7)^{1 / 3}}{2}$