Question

The moment of inertia of a solid sphere about an axis passing through its centre is $0.4{\text{kg-m}}^2$. If mass of the sphere is kept same, but its density is increased eight times, find the new moment of inertia of the sphere about the same axis.

• A. ${\text{0.4 kg-m}}^2$
• B. ${\text{0.1 kg-m}}^2$
• C. ${\text{0.3 kg-m}}^2$
• D. ${\text{0.2 kg-m}}^2$

Answer 1

The correct option is B ${\text{0.1 kg-m}}^2$

We know,
$\text{Density}\left(\rho \right)$$=\frac{\text{Mass}\left(M\right)}{\text{Volume}\left(V\right)}$
$\text{Volume}\left(V\right)=\frac{4\pi r^3}{3}$
If density is increased 8 times and mass is kept same,
New density, $\rho {}^{\prime }=8\times \rho$
$\Rightarrow \frac{M}{V^{\prime }}=8\times \frac{M}{V}$
$\Rightarrow V^{\prime }=\frac{V}{8}$
$\Rightarrow \frac{4\pi {r^{\prime }}^3}{3}=\frac{1}{8}\times \frac{4\pi r^3}{3}$
$\Rightarrow r^{\prime }=\frac{r}{2}$ (new radius)
We know moment of inertia of solid sphere about an axis passing through its $\text{C.O.M}$
$I=\frac{2}{5}M{r}^2$
New MOI, $I^{\prime }=\frac{2}{5}M{r^{\prime }}^2$

$\therefore \frac{I^{\prime }}{I}=\frac{{r^{\prime }}^2}{r^2}=\frac{1}{4}$

$\Rightarrow I^{\prime }=\frac{I}{4}=\frac{0.4}{4}=0.1{\text{kg-m}}^2$