A sphere of material of relative density 8 has a concentric spherical cavity and just sinks in water. If the radius of the sphere is 2cm , then the volume of the cavity is

Given,

The sphere's relative density is $d_B=8$

Radius of the sphere is R = 2cm

Relative density of water $d_W=1$

let r be the radius of the cavity,

$$\begin{gathered} \frac{4}{3}\mathrm{\pi }({\mathrm{R}}^3-{\mathrm{\rho }}^3){\mathrm{d}}_{\mathrm{B}}\mathrm{g}=\frac{4}{3}{\mathrm{\pi R}}^3{\mathrm{d}}_{\mathrm{w}}\mathrm{g} \\ \frac{8}{1}=\frac{{\mathrm{R}}^3}{{\mathrm{R}}^3-{\mathrm{r}}^3} \end{gathered}$$

$$\begin{gathered} 1-\frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}=\frac{1}{8} \\ \frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}=1-\frac{1}{8} \end{gathered}$$

$$\begin{gathered} \frac{{\mathrm{R}}^3}{{\mathrm{r}}^3}=\frac{7}{8} \\ {\mathrm{r}}^3=\frac{8}{7}{\mathrm{R}}^3 =\frac{7}{8}{\left(2\right)}^3=7 \end{gathered}$$

$\mathrm{Cavity} \mathrm{volume}=\frac{4}{3}{\mathrm{\pi r}}^3=\frac{4}{3}\times \frac{22}{7}\times 7=\frac{88}{3}{\mathrm{cm}}^3$