The moment of inertia of a uniform thin rod of length `L` and mass `M` about an axis passing through a point at a distance of `L//3` from one of its ends and perpendicular to the rod is

Options

A. `(7ML^(2))/48`
B. `(ML^(2))/6`
C. `(ML^(2))/9`
D. `(ML^(2))/3`

Explanation

To find the moment of inertia of a uniform thin rod of length \( L \) and mass \( M \) about an axis passing through a point at a distance of \( \frac{L}{3} \) from one of its ends and perpendicular to the rod, we can use the parallel axis theorem.

### Step-by-Step Solution:

1. **Identify the Moment of Inertia about the Center of Mass (CM)**:
The moment of inertia \( I_{CM} \) of a uniform thin rod about an axis passing through its center and perpendicular to its length is given by:
\[ I_{CM} = \frac{1}{12} ML^2 \]

2. **Determine the Distance from the Center of Mass to the New Axis**:
The center of mass of the rod is located at a distance of \( \frac{L}{2} \) from one end. The new axis is at a distance of \( \frac{L}{3} \) from one end. Therefore, the distance \( d \) from the center of mass to the new axis is:
\[ d = \left| \frac{L}{2} - \frac{L}{3} \right| \]
To find \( d \), we need a common denominator:
\[ d = \left| \frac{3L}{6} - \frac{2L}{6} \right| = \left| \frac{L}{6} \right| = \frac{L}{6} \]

3. **Apply the Parallel Axis Theorem**:
The parallel axis theorem states that:
\[ I = I_{CM} + Md^2 \]
Substituting the values we have:
\[ I = \frac{1}{12} ML^2 + M \left( \frac{L}{6} \right)^2 \]

4. **Calculate \( Md^2 \)**:
\[ Md^2 = M \left( \frac{L}{6} \right)^2 = M \cdot \frac{L^2}{36} = \frac{ML^2}{36} \]

5. **Combine the Moments of Inertia**:
Now substituting back into the equation:
\[ I = \frac{1}{12} ML^2 + \frac{ML^2}{36} \]
To add these fractions, we need a common denominator, which is 36:
\[ I = \frac{3ML^2}{36} + \frac{ML^2}{36} = \frac{4ML^2}{36} \]

6. **Simplify the Expression**:
\[ I = \frac{ML^2}{9} \]

### Final Answer:
The moment of inertia of the rod about the specified axis is:
\[ I = \frac{ML^2}{9} \]