Four identical solid spheres each of mass 'm' and radius 'a' are placed with their centres on the four corners of a square of side 'b'. The moment of inertia of the system about one side of square where the axis of rotation is parallel to the plane of the square is :

Options

A. `4/5ma^2+2mb^2`
B. `8/5ma^2+mb^2`
C. `8/5ma^2+2mb^2`
D. `4/5ma^2`

Answer

To find the moment of inertia of the system of four identical solid spheres about one side of the square, we can follow these steps:

### Step 1: Understand the Configuration
We have four identical solid spheres, each of mass \( m \) and radius \( a \), placed at the corners of a square with side length \( b \). The centers of the spheres coincide with the corners of the square.

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

### Step 3: Calculate the Moment of Inertia for Spheres A and B
For the two spheres located at the corners along the axis (let's say A and B), their moment of inertia about the side of the square can be calculated directly since the axis passes through their centers.

The moment of inertia for each sphere about the axis through their centers is:
\[ I_A = I_B = \frac{2}{5} m a^2 \]
Thus, the total moment of inertia for spheres A and B is:
\[ I_{AB} = I_A + I_B = \frac{2}{5} m a^2 + \frac{2}{5} m a^2 = \frac{4}{5} m a^2 \]

### Step 4: Calculate the Moment of Inertia for Spheres C and D
For spheres C and D, which are located at the opposite corners, we need to use the parallel axis theorem to find their contribution to the moment of inertia about the side of the square.

The distance from the center of each sphere to the axis (which is at a distance \( b \) from their centers) is \( b \). The moment of inertia about the axis through the center of each sphere is:
\[ I_{C} = I_{D} = \frac{2}{5} m a^2 \]
Using the parallel axis theorem:
\[ I' = I + md^2 \]
where \( d \) is the distance from the center of the sphere to the axis of rotation.

Thus, for spheres C and D:
\[ I_C' = \frac{2}{5} m a^2 + m b^2 \]
\[ I_D' = \frac{2}{5} m a^2 + m b^2 \]

### Step 5: Total Moment of Inertia for C and D
The total moment of inertia for spheres C and D is:
\[ I_{CD} = I_C' + I_D' = \left( \frac{2}{5} m a^2 + m b^2 \right) + \left( \frac{2}{5} m a^2 + m b^2 \right) = \frac{4}{5} m a^2 + 2 m b^2 \]

### Step 6: Total Moment of Inertia of the System
Now, we can find the total moment of inertia \( I \) of the system about one side of the square:
\[ I = I_{AB} + I_{CD} = \frac{4}{5} m a^2 + \left( \frac{4}{5} m a^2 + 2 m b^2 \right) \]
\[ I = \frac{4}{5} m a^2 + \frac{4}{5} m a^2 + 2 m b^2 = \frac{8}{5} m a^2 + 2 m b^2 \]

### Final Answer
The moment of inertia of the system about one side of the square is:
\[ I = \frac{8}{5} m a^2 + 2 m b^2 \]