Four spheres, each of mass M and radius r are situated at the four corners of square of side R . The moment of inertia of the system about an axis perpendicular to the plane of square and passing through its centre will be

The correct option is: 2M(4r² + 5 R²) / 5

To find the moment of inertia of the system about an axis perpendicular to the plane of the square and passing through its center, we can break down the calculation into the following steps:

Step 1: Moment of inertia of a single sphere about its own diameter
Each sphere has a mass $M$ and a radius $r$. Assuming the spheres are solid, the moment of inertia $I_{\mathrm{cm}}$ of a solid sphere of mass $M$ and radius $r$ about an axis passing through its center of mass is given by:
$I_{\mathrm{cm}}=\frac{2}{5}M{r}^2$

Step 2: Distance from the center of the square to the center of each sphere
The spheres are situated at the four corners of a square of side $R$. Let the corners be at a distance $d$ from the center of the square.
The length of the diagonal of the square of side $R$ is:
$\text{Diagonal}=\sqrt{2}R$
The distance $d$ from the center of the square to any corner is half of the diagonal length:
$d=\frac{\sqrt{2}R}{2}=\frac{R}{\sqrt{2}}$
Therefore, the square of the distance is:
$d^2=\frac{R^2}{2}$

Step 3: Moment of inertia of one sphere about the central axis using the parallel axis theorem
According to the parallel axis theorem, the moment of inertia $I_1$ of one sphere about an axis passing through the center of the square and perpendicular to its plane is:
$I_1=I_{\mathrm{cm}}+M{d}^2$
Substituting the values of $I_{\mathrm{cm}}$ and $d^2$:
$I_1=\frac{2}{5}M{r}^2+M\left(\frac{R^2}{2}\right)$

Step 4: Total moment of inertia of the system
Since there are four identical spheres symmetrically positioned at the four corners of the square, the total moment of inertia $I$ of the system is four times the moment of inertia of a single sphere:
$I=4I_1$
$I=4\left[\frac{2}{5}M{r}^2+\frac{1}{2}M{R}^2\right]$
To simplify this expression, we can factor out $\frac{M}{5}$ or rewrite it to match the options:
$I=4·\frac{M}{10}\left[4r^2+5R^2\right]$
$I=\frac{2M}{5}\left[4r^2+5R^2\right]$

Thus, the total moment of inertia of the system is $\frac{2M(4r^2+5R^2)}{5}$.