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 `60^(@)`. 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 find the moment of inertia of a bent rod, we can follow these steps:

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

### Step 2: Determine the Length and Mass of Each Half
When the rod is bent in the middle, it is divided into two halves. Each half has:
- Length = \( \frac{L}{2} \)
- Mass = \( \frac{M}{2} \)

### Step 3: Calculate the Moment of Inertia of Each Half
The moment of inertia of a rod about one end is given by:
\[ I_{\text{end}} = \frac{1}{3} m l^2 \]
For each half of the rod:
\[ I_1 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{3} \cdot \frac{M}{2} \cdot \frac{L^2}{4} = \frac{ML^2}{24} \]

### Step 4: Consider the Angle Between the Two Halves
When the two halves are bent to form an angle of \(60^\circ\), we need to account for the contribution of both halves to the total moment of inertia. Since the two halves are perpendicular to each other, we can use the parallel axis theorem and the perpendicular axis theorem.

### Step 5: Total Moment of Inertia Calculation
The total moment of inertia for the two halves can be calculated as:
\[ I_{\text{total}} = I_1 + I_2 \]
Since both halves have the same moment of inertia and are at an angle of \(60^\circ\):
\[ I_{\text{total}} = 2 \cdot I_1 \cdot \cos^2(30^\circ) \]
Using \( \cos(30^\circ) = \frac{\sqrt{3}}{2} \):
\[ I_{\text{total}} = 2 \cdot \frac{ML^2}{24} \cdot \left(\frac{\sqrt{3}}{2}\right)^2 = 2 \cdot \frac{ML^2}{24} \cdot \frac{3}{4} = \frac{3ML^2}{48} = \frac{ML^2}{16} \]

### Final Answer
Thus, the moment of inertia of the bent rod about the same axis is:
\[ I = \frac{ML^2}{16} \]