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. `2 ML^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 1: Identify the System
We have two masses, each of mass \( M \), attached to the ends of a rigid massless rod of length \( L \).

### Step 2: Determine the Position of the Center of Mass
The center of mass (COM) of the system can be found since the two masses are equal and symmetrically placed. The center of mass will be located at the midpoint of the rod, which is at a distance of \( \frac{L}{2} \) from each mass.

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

### Step 4: Calculate the Moment of Inertia for Each Mass
For each mass \( M \):
- The distance from the center of mass to each mass is \( r = \frac{L}{2} \).

Thus, the moment of inertia for each mass is:
\[ I_1 = M \left(\frac{L}{2}\right)^2 = M \frac{L^2}{4} \]
\[ I_2 = M \left(\frac{L}{2}\right)^2 = M \frac{L^2}{4} \]

### Step 5: Sum the Contributions to Find Total Moment of Inertia
Now, we sum the contributions from both masses:
\[ I = I_1 + I_2 = M \frac{L^2}{4} + M \frac{L^2}{4} = 2M \frac{L^2}{4} = \frac{ML^2}{2} \]

### Final Result
The moment of inertia of the system about an axis passing through the center of mass and perpendicular to its length is:
\[ I = \frac{ML^2}{2} \]