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. `(ML^2)/12`
B. `(ML^2)/9`
C. `(7ML^2)/48`
D. `(ML^2)/48`

Answer

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 follow these steps:

### Step 1: Identify the distances
We need to find the distance from the center of mass of the rod to the point where the axis is located. The center of mass of a uniform thin rod is located at its midpoint, which is at a distance of \( \frac{L}{2} \) from either end.

Given that the axis is at a distance of \( \frac{L}{3} \) from one end, we can calculate the distance from the center of mass to this point.

\[ \text{Distance from center of mass to point} = \frac{L}{2} - \frac{L}{3} \]

### Step 2: Calculate the distance
To perform the subtraction, we need a common denominator:

\[ \frac{L}{2} = \frac{3L}{6}, \quad \frac{L}{3} = \frac{2L}{6} \]

Now, subtract:

\[ \text{Distance} = \frac{3L}{6} - \frac{2L}{6} = \frac{L}{6} \]

### Step 3: Use the parallel axis theorem
The parallel axis theorem states:

\[ I = I_{cm} + Md^2 \]

Where:
- \( I \) is the moment of inertia about the new axis,
- \( I_{cm} \) is the moment of inertia about the center of mass,
- \( M \) is the mass of the rod,
- \( d \) is the distance from the center of mass to the new axis.

For a thin rod about its center of mass:

\[ I_{cm} = \frac{ML^2}{12} \]

Now, substituting into the parallel axis theorem:

\[ I = \frac{ML^2}{12} + M\left(\frac{L}{6}\right)^2 \]

### Step 4: Calculate \( Md^2 \)
Calculating \( Md^2 \):

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

### Step 5: Combine the terms
Now, substituting back into the equation for \( I \):

\[ I = \frac{ML^2}{12} + \frac{ML^2}{36} \]

To combine these fractions, we need a common denominator, which is 36:

\[ \frac{ML^2}{12} = \frac{3ML^2}{36} \]

Now, adding the two fractions:

\[ I = \frac{3ML^2}{36} + \frac{ML^2}{36} = \frac{4ML^2}{36} = \frac{ML^2}{9} \]

### Conclusion
Thus, the moment of inertia of the rod about the specified axis is:

\[ \boxed{\frac{ML^2}{9}} \]