A thin wire of mass m and length l is bent in the form of a ring . Moment of inertia of that ring about an axis passing through its centre and perpendicular to its through its centre and perpendicular to its plane is

Options

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

Explanation

To find the moment of inertia of a thin wire bent in the form of a ring about an axis passing through its center and perpendicular to its plane, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Parameters**:
- Let the mass of the wire be \( m \).
- Let the length of the wire be \( l \).

2. **Relate Length to Radius**:
- The wire is bent into the shape of a ring. The circumference of the ring is equal to the length of the wire.
- Thus, we have the equation for the circumference of the ring:
\[ l = 2\pi r \]
- From this, we can solve for the radius \( r \):
\[ r = \frac{l}{2\pi} \]

3. **Use the Formula for Moment of Inertia**:
- The formula for the moment of inertia \( I \) of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I = m r^2 \]

4. **Substitute the Radius**:
- Now, substitute the expression for \( r \) into the moment of inertia formula:
\[ I = m \left(\frac{l}{2\pi}\right)^2 \]

5. **Simplify the Expression**:
- Expanding this, we get:
\[ I = m \cdot \frac{l^2}{(2\pi)^2} \]
- This simplifies to:
\[ I = \frac{m l^2}{4\pi^2} \]

6. **Final Result**:
- Therefore, the moment of inertia of the ring about the specified axis is:
\[ I = \frac{m l^2}{4\pi^2} \]