The moment of inertia of a rod about an axis through its centre and perpendicular to it, is `(1)/(12)ML^(2)` (where, M is the mass and L is length of the rod). The rod is bent in the middle, so that two halves make an angle of `45^(@)`. The moment of inertia of the bent rod about the same axis would be

Options

A. `(1)/(48)ML^(2)`
B. `(1)/(12)ML^(2)`
C. `(1)/(24)ML^(2)`
D. `(ML^(2))/(8sqrt(3))`

Explanation

To solve the problem, we need to find the moment of inertia of a bent rod about an axis through its center and perpendicular to it after it has been bent at an angle of 45 degrees.

### Step-by-step Solution:

1. **Understand the Initial Moment of Inertia**:
The moment of inertia (I) of a straight rod about an axis through its center and perpendicular to its length is given by:
\[ I = \frac{1}{12} ML^2 \]
where \( M \) is the mass of the rod and \( L \) is its length.

2. **Bending the Rod**:
When the rod is bent in the middle to form an angle of \( 45^\circ \), it effectively creates two segments (each of length \( \frac{L}{2} \)) that are at an angle to each other.

3. **Calculate Moment of Inertia for Each Segment**:
Each half of the rod has a mass of \( \frac{M}{2} \) and a length of \( \frac{L}{2} \). The moment of inertia of each half about its own center (which is at the midpoint of the half-rod) is:
\[ I_{half} = \frac{1}{12} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{12} \cdot \frac{M}{2} \cdot \frac{L^2}{4} = \frac{ML^2}{96} \]

4. **Use the Parallel Axis Theorem**:
To find the moment of inertia about the original axis (the axis through the center of the entire rod), we need to apply the parallel axis theorem. The distance from the center of each half-rod to the axis is \( \frac{L}{4} \) (since the center of the bent rod is at the midpoint of the original rod).

The moment of inertia of each half-rod about the original axis is given by:
\[ I_{total} = I_{half} + \frac{M}{2} \left(\frac{L}{4}\right)^2 \]
where \( \frac{M}{2} \) is the mass of each half-rod and \( \left(\frac{L}{4}\right)^2 \) is the square of the distance from the center of the half-rod to the original axis.

Substituting the values:
\[ I_{total} = \frac{ML^2}{96} + \frac{M}{2} \cdot \frac{L^2}{16} \]
Simplifying the second term:
\[ \frac{M}{2} \cdot \frac{L^2}{16} = \frac{ML^2}{32} \]

5. **Combine the Moments of Inertia**:
Now, we need to combine both contributions:
\[ I_{total} = \frac{ML^2}{96} + \frac{ML^2}{32} \]
To add these fractions, we need a common denominator, which is 96:
\[ \frac{ML^2}{32} = \frac{3ML^2}{96} \]
Therefore:
\[ I_{total} = \frac{ML^2}{96} + \frac{3ML^2}{96} = \frac{4ML^2}{96} = \frac{ML^2}{24} \]

6. **Final Result**:
The moment of inertia of the bent rod about the same axis is:
\[ I = \frac{ML^2}{24} \]