A thin rod of mass `M` and length `L` is bent into a circular ring. The expression for moment of inertia of ring about an axis passing through its diameter is

Options

A. `(ML^(2))/(2pi^(2))`
B. `(ML^(2))/(4 pi^(2))`
C. `(ML^(2))/(8pi^(2))`
D. `(ML^(2))/(pi^(2))`

Explanation

To find the moment of inertia of a thin rod bent into a circular ring about an axis passing through its diameter, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have a thin rod of mass \( M \) and length \( L \). When bent into a circular ring, the length of the rod becomes the circumference of the ring.

2. **Finding the Radius of the Ring**:
The circumference \( C \) of the ring can be expressed as:
\[ C = 2\pi r \]
Since the circumference is equal to the length of the rod, we have:
\[ 2\pi r = L \]
From this, we can solve for the radius \( r \):
\[ r = \frac{L}{2\pi} \]

3. **Using the Moment of Inertia Formula**:
The moment of inertia \( I \) of a circular ring about an axis passing through its diameter is given by the formula:
\[ I = \frac{1}{2} M r^2 \]

4. **Substituting the Radius**:
Now, we substitute the expression for \( r \) into the moment of inertia formula:
\[ I = \frac{1}{2} M \left(\frac{L}{2\pi}\right)^2 \]

5. **Calculating the Moment of Inertia**:
Simplifying the expression:
\[ I = \frac{1}{2} M \cdot \frac{L^2}{(2\pi)^2} \]
\[ I = \frac{M L^2}{2 \cdot 4\pi^2} \]
\[ I = \frac{M L^2}{8\pi^2} \]

6. **Final Expression**:
Therefore, the moment of inertia of the ring about an axis passing through its diameter is:
\[ I = \frac{M L^2}{8\pi^2} \]