Physics - Rotation Question with Solution | TestHub

Let the mass of the original solid sphere be $M$ and its radius be $R$.

The moment of inertia of the solid sphere about its diameter is $I=\frac{2}{5}M{R}^2$.

Let the mass of each small sphere be $m$ and its radius be $r$.

Since the original sphere is cast into $27$ small spheres, the total volume remains constant:

$\frac{4}{3}\pi R^3=27\times \frac{4}{3}\pi r^3$

$R^3=27r^3\Rightarrow r=\frac{R}{3}$

Also, the mass of each small sphere is $m=\frac{M}{27}$.

The moment of inertia of each small sphere about its diameter is:

$I^{\prime }=\frac{2}{5}m{r}^2$

Substituting the values of $m$ and $r$:

$I^{\prime }=\frac{2}{5}\left(\frac{M}{27}\right){\left(\frac{R}{3}\right)}^2$

$I^{\prime }=\frac{2}{5}\frac{M}{27}\frac{R^2}{9}$

$I^{\prime }=\frac{1}{243}\left(\frac{2}{5}M{R}^2\right)$

$I^{\prime }=\frac{I}{243}$

Answer: $\frac{\mathrm{I}}{243}$