The moment of inertia of a uniform rod about a perpendicular axis passing through one end is `I_(1)`. The same rod is bent into a ring and its moment of inertia about a diameter is `I_(2)`. Then `I_(1)//I_(2)` is

Options

A. `8/3π^2`
B. `5/3π^2`
C. `2/3π^2`
D. `4/3π^2`

Answer

To solve the problem, we need to find the ratio of the moment of inertia of a uniform rod about a perpendicular axis passing through one end (I1) to the moment of inertia of the same rod bent into a ring about a diameter (I2).

### Step-by-Step Solution:

1. **Calculate I1 (Moment of Inertia of the Rod)**:
- The moment of inertia of a uniform rod about an axis perpendicular to it and passing through one end is given by the formula:
\[ I_1 = \frac{1}{3} m L^2 \]
- Here, \(m\) is the mass of the rod and \(L\) is the length of the rod.

2. **Bend the Rod into a Ring**:
- When the rod is bent into a ring, the circumference of the ring will equal the length of the rod:
\[ 2\pi r = L \quad \Rightarrow \quad r = \frac{L}{2\pi} \]
- Here, \(r\) is the radius of the ring.

3. **Calculate I2 (Moment of Inertia of the Ring)**:
- The moment of inertia of a ring about a diameter is given by:
\[ I_2 = \frac{1}{2} m r^2 \]
- Substituting the value of \(r\) from the previous step:
\[ I_2 = \frac{1}{2} m \left(\frac{L}{2\pi}\right)^2 = \frac{1}{2} m \frac{L^2}{4\pi^2} = \frac{m L^2}{8\pi^2} \]

4. **Find the Ratio I1/I2**:
- Now, we can find the ratio of \(I_1\) to \(I_2\):
\[ \frac{I_1}{I_2} = \frac{\frac{1}{3} m L^2}{\frac{m L^2}{8\pi^2}} = \frac{\frac{1}{3}}{\frac{1}{8\pi^2}} = \frac{8\pi^2}{3} \]

5. **Final Result**:
- Therefore, the ratio \( \frac{I_1}{I_2} \) is:
\[ \frac{I_1}{I_2} = \frac{8\pi^2}{3} \]