Two masses each of mass `M` are attached to the end of a rigid massless rod of length `L`. The moment of interia of the system about an axis passing centre of mass and perpendicular to its length is.

Options

A. `ML^(2)`/4
B. `ML^(2)`/2
C. `ML^(2)`
D. `2ML^(2)`

Explanation

To find the moment of inertia of the system about an axis passing through the center of mass and perpendicular to the length of the rod, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the System**: We have two masses, each of mass \( M \), attached to the ends of a massless rod of length \( L \). The center of mass of the system will be located at the midpoint of the rod.

2. **Determine the Distance from the Axis**: Since the rod is of length \( L \), the distance from the center of mass to each mass is \( \frac{L}{2} \).

3. **Formula for Moment of Inertia**: The moment of inertia \( I \) about an axis is given by the formula:
\[ I = \sum m_i r_i^2 \]
where \( m_i \) is the mass and \( r_i \) is the distance from the axis of rotation.

4. **Calculate Moment of Inertia for Each Mass**:
- For the first mass \( M \):
\[ I_1 = M \left(\frac{L}{2}\right)^2 = M \frac{L^2}{4} \]
- For the second mass \( M \):
\[ I_2 = M \left(\frac{L}{2}\right)^2 = M \frac{L^2}{4} \]

5. **Total Moment of Inertia**: The total moment of inertia \( I \) for the system is the sum of the moments of inertia of both masses:
\[ I = I_1 + I_2 = M \frac{L^2}{4} + M \frac{L^2}{4} = 2 \left(M \frac{L^2}{4}\right) = M \frac{L^2}{2} \]

6. **Final Result**: Therefore, the moment of inertia of the system about the specified axis is:
\[ I = \frac{ML^2}{2} \]

### Conclusion:
The moment of inertia of the system about an axis passing through the center of mass and perpendicular to its length is \( \frac{ML^2}{2} \).

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