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. `(pi^(2))/(3)`
B. `(2pi^(2))/(3)`
C. `(4pi^(2))/(3)`
D. `(8pi^(2))/(3)`

Explanation

To solve the problem, we need to find the ratio \( \frac{I_1}{I_2} \) where \( I_1 \) is the moment of inertia of a uniform rod about a perpendicular axis passing through one end, and \( I_2 \) is the moment of inertia of the same rod bent into a ring about a diameter.

### Step 1: Calculate \( I_1 \)

The moment of inertia \( I_1 \) of a uniform rod of mass \( m \) and length \( l \) about an axis perpendicular to it and passing through one end is given by the formula:

\[
I_1 = \frac{1}{3} m l^2
...