Four spheres of diameter 2a and mass M are placed with their centres on the four corners of a square of side b. Then moment of inertia of the system about an axis about one of the sides of the square is :-

Options

A. `Ma^(2)+2Mb^(2)`
B. `Ma^(2)`
C. `Ma^(2)+4Mb^(2)`
D. `(8)/(5)Ma^(2)+2Mb^(2)`

Answer

To find the moment of inertia of the system of four spheres placed at the corners of a square about one of its sides, we can follow these steps:

### Step 1: Understand the Configuration
We have four spheres, each with a diameter of 2a (thus a radius of a), and mass M. They are positioned at the corners of a square with side length b.

### Step 2: Moment of Inertia of a Single Sphere
The moment of inertia (I) of a solid sphere about an axis through its center is given by the formula:
\[ I_{\text{center}} = \frac{2}{5} M a^2 \]

### Step 3: Use the Parallel Axis Theorem
To find the moment of inertia about an axis that is not through the center of the sphere, we use the Parallel Axis Theorem:
\[ I = I_{\text{center}} + Md^2 \]
where \(d\) is the distance from the center of the sphere to the new axis.

### Step 4: Calculate the Moment of Inertia for Each Sphere
1. **Sphere at Corner A**: The distance from the center of sphere A to the side AB is \(d = 0\) (since it lies on the axis).
\[ I_A = \frac{2}{5} M a^2 + M(0)^2 = \frac{2}{5} M a^2 \]

2. **Sphere at Corner B**: The distance from the center of sphere B to the side AB is \(d = b\).
\[ I_B = \frac{2}{5} M a^2 + M(b)^2 = \frac{2}{5} M a^2 + M b^2 \]

3. **Sphere at Corner C**: The distance from the center of sphere C to the side AB is \(d = b\) (same as sphere B).
\[ I_C = \frac{2}{5} M a^2 + M(b)^2 = \frac{2}{5} M a^2 + M b^2 \]

4. **Sphere at Corner D**: The distance from the center of sphere D to the side AB is \(d = 0\) (same as sphere A).
\[ I_D = \frac{2}{5} M a^2 + M(0)^2 = \frac{2}{5} M a^2 \]

### Step 5: Total Moment of Inertia
Now, we can sum the moments of inertia of all four spheres:
\[ I_{\text{total}} = I_A + I_B + I_C + I_D \]
Substituting the values we calculated:
\[ I_{\text{total}} = \left(\frac{2}{5} M a^2\right) + \left(\frac{2}{5} M a^2 + M b^2\right) + \left(\frac{2}{5} M a^2 + M b^2\right) + \left(\frac{2}{5} M a^2\right) \]
\[ I_{\text{total}} = 4 \left(\frac{2}{5} M a^2\right) + 2 M b^2 \]
\[ I_{\text{total}} = \frac{8}{5} M a^2 + 2 M b^2 \]

### Final Answer
Thus, the moment of inertia of the system about one of the sides of the square is:
\[ I_{\text{total}} = \frac{8}{5} M a^2 + 2 M b^2 \]