Moment of inertia of a straight wire about an axis perpendicular to the wire passing through one of its end is I.
This wire is now framed into a circle (a ring) of single turn. The moment of inertia of this ring about an axis passing through centre and perpendicular to its plane would be

Options

A. `((3)/(pi^(2)))l`
B. `((3)/(4pi^(2)))l`
C. `((pi^(2))/(3))l`
D. `((4pi^(2))/(3))l`

Answer

To solve the problem, we need to find the moment of inertia of a ring formed by bending a straight wire, given that the moment of inertia of the straight wire about an axis perpendicular to it and passing through one of its ends is \( I \).

### Step-by-Step Solution:

1. **Understand the Moment of Inertia of the Straight Wire:**
- The moment of inertia \( I \) of the straight wire about an axis perpendicular to it and passing through one of its ends can be expressed as:
\[ I = \frac{ML^2}{3} \]
where \( M \) is the mass of the wire and \( L \) is its length.

2. **Convert the Straight Wire into a Ring:**
- When the straight wire is bent into a ring, the length of the wire remains the same. The circumference of the ring is given by:
\[ C = 2\pi r \]
where \( r \) is the radius of the ring. Setting this equal to the length of the wire:
\[ 2\pi r = L \]
From this, we can express the radius \( r \) in terms of the length \( L \):
\[ r = \frac{L}{2\pi} \]

3. **Find the Moment of Inertia of the Ring:**
- The moment of inertia \( I' \) of a ring about an axis passing through its center and perpendicular to its plane is given by:
\[ I' = m r^2 \]
where \( m \) is the mass of the ring. Since the mass of the wire remains the same, we can use \( M \) for \( m \):
\[ I' = M r^2 \]
Substituting the expression for \( r \):
\[ I' = M \left(\frac{L}{2\pi}\right)^2 = M \frac{L^2}{4\pi^2} \]

4. **Relate \( ML^2 \) to \( I \):**
- From the moment of inertia of the straight wire, we have:
\[ ML^2 = 3I \]
Now, substitute this into the equation for \( I' \):
\[ I' = \frac{3I}{4\pi^2} \]

5. **Final Result:**
- Therefore, the moment of inertia of the ring about an axis passing through its center and perpendicular to its plane is:
\[ I' = \frac{3I}{4\pi^2} \]

### Conclusion:
The moment of inertia of the ring is \( \frac{3I}{4\pi^2} \).