Moment of inertia of a thin uniform rod rotating about the perpendicular axis passing through its center is l. If the same rod is bent into a ring and its moment of inertia about its diameter is I. then the ratio `(I)/(I)` is

Options

A. `(8)/(3)pi^(2)`
B. `(5)/(3)pi^(2)`
C. `(3)/(2)pi^(2)`
D. `(2)/(3)pi^(2)`

Explanation

To solve the problem, we need to find the ratio of the moment of inertia of a thin uniform rod (I) rotating about a perpendicular axis passing through its center to the moment of inertia of the same rod bent into a ring (I') about its diameter.

### Step-by-Step Solution:

1. **Determine the Moment of Inertia of the Rod (I)**:
The moment of inertia (I) of a thin uniform rod of mass \( m \) and length \( l \) about an axis perpendicular to its length and passing through its center is given by the formula:
\[ I = \frac{1}{12} m l^2 \]

2. **Bend the Rod into a Ring**:
When the rod is bent into a ring, the mass remains the same, but we need to determine the radius of the ring. The circumference of the ring is equal to the length of the rod:
\[ 2\pi r = l \quad \Rightarrow \quad r = \frac{l}{2\pi} \]

3. **Determine the Moment of Inertia of the Ring (I')**:
The moment of inertia (I') of a ring about an axis through its diameter is given by:
\[ I' = \frac{1}{2} m r^2 \]
Substituting the expression for \( r \):
\[ I' = \frac{1}{2} m \left(\frac{l}{2\pi}\right)^2 = \frac{1}{2} m \frac{l^2}{4\pi^2} = \frac{m l^2}{8\pi^2} \]

4. **Find the Ratio of I to I'**:
Now we can find the ratio of the moment of inertia of the rod to that of the ring:
\[ \frac{I}{I'} = \frac{\frac{1}{12} m l^2}{\frac{m l^2}{8\pi^2}} \]
Simplifying this expression:
\[ \frac{I}{I'} = \frac{1}{12} \cdot \frac{8\pi^2}{1} = \frac{8\pi^2}{12} = \frac{2\pi^2}{3} \]

5. **Final Result**:
Thus, the ratio of the moment of inertia of the rod to the moment of inertia of the ring about its diameter is:
\[ \frac{I}{I'} = \frac{2\pi^2}{3} \]

### Conclusion:
The final answer is \( \frac{2\pi^2}{3} \).