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))/(m_(1)+m_(2)) l^(2)`
B. `(m_(1)+m_(2))/(m_(1)m_(2)) l^(2)`
C. `(m_(1)+m_(2))l^(2)`
D. `sqrt(m_(1)m_(2))l^(2)`

Answer

To find the moment of inertia of a system consisting of a light rod of length \( l \) with two masses \( m_1 \) and \( m_2 \) attached to its ends 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 (massless) rod of length \( l \) with two point masses \( m_1 \) and \( m_2 \) attached at its ends.

2. **Determine the Position of the Center of Mass (CM)**:
The center of mass \( x_{cm} \) for two point masses can be calculated using the formula:
\[ x_{cm} = \frac{m_1 \cdot x_1 + m_2 \cdot x_2}{m_1 + m_2} \]
where \( x_1 \) and \( x_2 \) are the distances of \( m_1 \) and \( m_2 \) from the center of mass.

3. **Set the Length of the Rod**:
Let the distance from \( m_1 \) to the center of mass be \( x_1 \) and from \( m_2 \) be \( x_2 \). Since the total length of the rod is \( l \):
\[ x_1 + x_2 = l \]

4. **Express Distances in Terms of Masses**:
From the center of mass formula, we can express \( x_1 \) and \( x_2 \):
\[ x_1 = \frac{m_2}{m_1 + m_2} \cdot l \]
\[ x_2 = \frac{m_1}{m_1 + m_2} \cdot l \]

5. **Calculate the Moment of Inertia (I)**:
The moment of inertia about the center of mass is given by:
\[ I_{cm} = m_1 \cdot x_1^2 + m_2 \cdot x_2^2 \]
Substituting \( x_1 \) and \( x_2 \):
\[ I_{cm} = m_1 \left(\frac{m_2}{m_1 + m_2} \cdot l\right)^2 + m_2 \left(\frac{m_1}{m_1 + m_2} \cdot l\right)^2 \]

6. **Simplify the Expression**:
\[ I_{cm} = m_1 \cdot \frac{m_2^2 l^2}{(m_1 + m_2)^2} + m_2 \cdot \frac{m_1^2 l^2}{(m_1 + m_2)^2} \]
\[ I_{cm} = \frac{l^2}{(m_1 + m_2)^2} \left( m_1 m_2^2 + m_2 m_1^2 \right) \]
\[ = \frac{l^2 m_1 m_2 (m_1 + m_2)}{(m_1 + m_2)^2} \]
\[ = \frac{m_1 m_2 l^2}{m_1 + m_2} \]

### Final Answer:
The moment of inertia of the system about the axis perpendicular to the rod and passing through the center of mass is:
\[ I_{cm} = \frac{m_1 m_2 l^2}{m_1 + m_2} \]