A thin uniform rod of mass `M` and Length `L` has its moment of inertia `I_(1)` about its perpendicular bisector. The rod is bend in the form of a semocircular arc. Now its moment of inertia through the centre of the semi circular arc and perpendicular to its plane is `I_(2)`. The ratio of `I_(2) : I_(2)` will be ____.

Options

A. `lt 1`
B. ` lt 1`
C. `= 1`
D. can't be said

Explanation

To solve the problem, we need to find the moments of inertia \( I_1 \) and \( I_2 \) and then calculate the ratio \( I_1 : I_2 \).

### Step-by-Step Solution:

1. **Calculate \( I_1 \) for the thin uniform rod about its perpendicular bisector:**
The moment of inertia \( I_1 \) of a thin uniform rod of mass \( M \) and length \( L \) about its perpendicular bisector is given by the formula:
\[ I_1 = \frac{1}{12} M L^2 \]

2. **Understanding the bending of the rod:**
When the rod is bent into a semicircular arc, the length of the arc remains \( L \). The radius \( R \) of the semicircular arc can be expressed in terms of \( L \):
\[ L = \pi R \implies R = \frac{L}{\pi} \]

3. **Calculate \( I_2 \) for the semicircular arc:**
The moment of inertia \( I_2 \) of a semicircular arc about its center and perpendicular to its plane can be calculated using the formula:
\[ I_2 = \frac{1}{2} M R^2 \]
Substituting \( R = \frac{L}{\pi} \):
\[ I_2 = \frac{1}{2} M \left(\frac{L}{\pi}\right)^2 = \frac{1}{2} M \frac{L^2}{\pi^2} = \frac{M L^2}{2 \pi^2} \]

4. **Calculate the ratio \( I_1 : I_2 \):**
Now we can find the ratio \( I_1 : I_2 \):
\[ \frac{I_1}{I_2} = \frac{\frac{1}{12} M L^2}{\frac{M L^2}{2 \pi^2}} = \frac{1}{12} \cdot \frac{2 \pi^2}{1} = \frac{\pi^2}{6} \]
Therefore, the ratio \( I_1 : I_2 \) is:
\[ I_1 : I_2 = \frac{1}{12} : \frac{1}{2 \pi^2} = \frac{\pi^2}{6} : 1 \]

### Final Answer:
The ratio \( I_1 : I_2 \) is \( \frac{\pi^2}{6} : 1 \).