M.I of a solid sphere about its diameter is $64kg{m}^2$ . If that sphere is recast into 8 identical small spheres, then M.I. of such small sphere about its diameter is

$\frac{2}{5}M{R}^2=64---\left(1\right)$ 

$$\begin{gathered} V_{big}=n{V}_{small} \\ \frac{4}{3}{\mathrm{\pi R}}^3=\mathrm{n}\frac{4}{3}{\mathrm{\pi r}}^3 \\ \mathrm{r}=\frac{\mathrm{R}}{2} \\ \mathrm{m}\propto {\mathrm{R}}^3\left(\mathrm{m}=\mathrm{V\rho }\right) \\ \frac{{\mathrm{m}}_2}{{\mathrm{m}}_1}={\left(\frac{{\mathrm{R}}_2}{{\mathrm{R}}_1}\right)}^3={\left(\frac{{\displaystyle \frac{{\mathrm{R}}_1}{2}}}{{\mathrm{R}}_1}\right)}^3=\frac{1}{8} \\ {\mathrm{m}}_2=\frac{\mathrm{M}}{8} \end{gathered}$$

              For small spheres $m=\frac{M}{8},r=\frac{R}{2}$

          $I\text{'}=\frac{2}{5}m{r}^2=\frac{2}{5}\times \frac{M}{8}\times {\left(\frac{R}{2}\right)}^2$

                = $\frac{1}{32}\left(\frac{2}{5}M{R}^2\right)=2kg{m}^2$