Question

Four solid spheres each of diameter $\sqrt{5}\text{cm}$ and mass $0.5\text{kg}$ are placed with their centres at the corners of a square of side $4\text{cm}$. The moment of inertia of the system about the diagonal of the square is $N\times {10}^{-4}{\text{kg m}}^2$, then $N$ is

• A. $7$
• B. $8$
• C. $9$
• D. $10$

Answer 1

The correct option is C $9$

Given that,
Diameter of the sphere, $D=\sqrt{5}\text{cm}$
Radius of the sphere, $r=\frac{\sqrt{5}}{2}$
Mass of the sphere, $m=0.5\text{kg}$

Let the moment of inertia of the sphere $A,B,C$ and $D$ is $I_A,I_B,I_C$ and $I_D$ respectively about $X-X^{\prime }$ axis.
The moment of inertia of each sphere about an axis passing through its centre is $\frac{2}{5}m{r}^2$

The moment of inertia of sphere $B$ and sphere $D$ about $X-X^{{}^{\prime }}$ is
$I_B=I_D=\frac{2}{5}m{r}^2$
Using parallel axes theorem, the moment of inertia of sphere $A$ and sphere $C$ about $X-X^{{}^{\prime }}$ is
$I_A=I_C=\frac{2}{5}m{r}^2+m{d}^2$
The moment of inertia of the system about the diagonal is
$I=I_A+I_B+I_C+I_D$
$\Rightarrow I=\frac{8}{5}m{r}^2+2m{d}^2$
Substituting the values
$d=\frac{a}{\sqrt{2}}=\frac{4}{\sqrt{2}}\text{cm}$
$\Rightarrow I=\frac{8}{5}\times 0.5\times (\frac{\sqrt{5}}{2}\times {10}^{-2}{)}^2+2\times 0.5\times (\frac{4}{\sqrt{2}}\times {10}^{-2}{)}^2$
$I=9\times {10}^{-4}{\text{kg-m}}^2$

Hence, option (c) is correct.