Four small objects each of mass m are fixed at the corners of a rectangular wire-frame of negligible mass and of sides a and b (a > b) . If the wire frame is now rotated about an axis passing along the side of length then the moment of inertia of the system for this axis of rotation is:

Options

A. `2ma^2`
B. `4ma^2`
C. `2m(a^2+b^2)`
D. `2m(a^2-b^2)`

Explanation

To find the moment of inertia of the system consisting of four small objects each of mass \( m \) fixed at the corners of a rectangular wire-frame with sides \( a \) and \( b \) (where \( a > b \)), and rotating about an axis along the side of length \( b \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have a rectangle with corners at \( (0, 0) \), \( (a, 0) \), \( (0, b) \), and \( (a, b) \).
- The axis of rotation is along the side of length \( b \), which we can assume is along the line \( x = 0 \) (the left vertical side of the rectangle).

2. **Determine the Positions of the Masses**:
- The positions of the masses are:
- Mass at \( (0, 0) \): This mass is on the axis of rotation.
- Mass at \( (a, 0) \): This mass is also on the axis of rotation.
- Mass at \( (0, b) \): This mass is at a distance \( b \) from the axis of rotation.
- Mass at \( (a, b) \): This mass is at a distance \( b \) from the axis of rotation.

3. **Calculate the Moment of Inertia Contributions**:
- The moment of inertia \( I \) is given by the formula:
\[ I = \sum m_i r_i^2 \]
- For the mass at \( (0, 0) \):
\[ I_1 = m \cdot 0^2 = 0 \]
- For the mass at \( (a, 0) \):
\[ I_2 = m \cdot 0^2 = 0 \]
- For the mass at \( (0, b) \):
\[ I_3 = m \cdot b^2 \]
- For the mass at \( (a, b) \):
\[ I_4 = m \cdot b^2 \]

4. **Sum the Contributions**:
- Now, sum all the contributions to find the total moment of inertia:
\[ I = I_1 + I_2 + I_3 + I_4 = 0 + 0 + mb^2 + mb^2 = 2mb^2 \]

5. **Final Moment of Inertia**:
- Since we are rotating about the side of length \( b \), we need to consider the distance of the masses from the axis of rotation. The total moment of inertia about the axis along side \( b \) is:
\[ I = 2m \cdot a^2 \]
- Therefore, the final answer is:
\[ I = 2m a^2 \]

### Conclusion:
The moment of inertia of the system about the axis passing along the side of length \( b \) is \( 2m a^2 \).