A light rod of length `l` has two masses `m_1` and `m_2` attached to its two ends. The moment of inertia of the system about an axis perpendicular to the rod and passing through the centre of mass is.

Options

A. `(m_1 + m_2) l^2`
B. `sqrt(m_1 m_2) l^2`
C. `(m_1 m_2)/(m_1 + m_2) l^2`
D. `(m_1 + m_2)/(m_1 m_2) l^2`

Explanation

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

### Step-by-Step Solution:

1. **Identify the System**: We have a light rod of length \( l \) with two masses \( m_1 \) and \( m_2 \) attached at its ends.

2. **Determine the Center of Mass (CM)**: The position of the center of mass \( R_{CM} \) of the system can be calculated using the formula:
\[ R_{CM} = \frac{m_1 \cdot 0 + m_2 \cdot l}{m_1 + m_2} = \frac{m_2 \cdot l}{m_1 + m_2} \]
Here, we consider the left end of the rod (where \( m_1 \) is attached) as the origin.

3. **Calculate Distances from CM**:
- The distance from \( m_1 \) to the center of mass \( R_1 \) is:
\[ R_1 = R_{CM} = \frac{m_2 \cdot l}{m_1 + m_2} \]
- The distance from \( m_2 \) to the center of mass \( R_2 \) is:
\[ R_2 = l - R_{CM} = l - \frac{m_2 \cdot l}{m_1 + m_2} = \frac{m_1 \cdot l}{m_1 + m_2} \]

4. **Apply the Moment of Inertia Formula**: The moment of inertia \( I \) about the center of mass is given by:
\[ I = m_1 R_1^2 + m_2 R_2^2 \]

5. **Substituting the Distances**:
- Substitute \( R_1 \) and \( R_2 \):
\[ I = m_1 \left(\frac{m_2 \cdot l}{m_1 + m_2}\right)^2 + m_2 \left(\frac{m_1 \cdot l}{m_1 + m_2}\right)^2 \]

6. **Simplifying the Expression**:
- Expanding the squares:
\[ I = m_1 \cdot \frac{m_2^2 \cdot l^2}{(m_1 + m_2)^2} + m_2 \cdot \frac{m_1^2 \cdot l^2}{(m_1 + m_2)^2} \]
- Combine the terms:
\[ I = \frac{l^2}{(m_1 + m_2)^2} (m_1 m_2^2 + m_2 m_1^2) = \frac{l^2 m_1 m_2 (m_1 + m_2)}{(m_1 + m_2)^2} \]

7. **Final Result**:
- Thus, the moment of inertia about the center of mass is:
\[ I = \frac{m_1 m_2 l^2}{m_1 + m_2} \]