A thin rod of length `L` and mass `M` is bent at its midpoint into two halves so that the angle between them is `90^@`. The moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod is.

Options

A. `(ML^2)/(24)`
B. `(ML^2)/(12)`
C. `(ML^2)/(6)`
D. `(sqrt(2) ML^2)/(24)`

Explanation

To find the moment of inertia of the bent rod about an axis passing through the bending point and perpendicular to the plane defined by the two halves of the rod, we can follow these steps:

### Step 1: Understand the Geometry of the Rod
The rod of length \( L \) is bent at its midpoint, creating two segments each of length \( \frac{L}{2} \) at a right angle (90 degrees) to each other.

### Step 2: Determine the Mass Distribution
Since the rod has a total mass \( M \), when bent, each half will have a mass of \( \frac{M}{2} \).

### Step 3: Calculate the Moment of Inertia for Each Half
The moment of inertia \( I \) of a thin rod about an axis through one end and perpendicular to its length is given by the formula:
\[ I = \frac{1}{3} m L^2 \]
For each half of the rod:
- Length of each half = \( \frac{L}{2} \)
- Mass of each half = \( \frac{M}{2} \)

Thus, the moment of inertia for one half about its own end (which is at the bending point) is:
\[ I_1 = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{1}{3} \cdot \frac{M}{2} \cdot \frac{L^2}{4} = \frac{ML^2}{24} \]

### Step 4: Calculate the Moment of Inertia for the Other Half
The second half of the rod is also \( \frac{L}{2} \) long and has the same mass \( \frac{M}{2} \). However, its moment of inertia needs to be calculated about the bending point, which is at a distance \( \frac{L}{2} \) from the axis of rotation. We will use the parallel axis theorem:
\[ I = I_{cm} + md^2 \]
Where:
- \( I_{cm} = \frac{1}{3} \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{ML^2}{24} \) (moment of inertia about its center of mass)
- \( d = \frac{L}{2} \) (distance from the center of mass to the bending point)

Calculating the moment of inertia for the second half:
\[ I_2 = \frac{ML^2}{24} + \left(\frac{M}{2}\right) \left(\frac{L}{2}\right)^2 = \frac{ML^2}{24} + \frac{M}{2} \cdot \frac{L^2}{4} = \frac{ML^2}{24} + \frac{ML^2}{8} = \frac{ML^2}{24} + \frac{3ML^2}{24} = \frac{4ML^2}{24} = \frac{ML^2}{6} \]

### Step 5: Total Moment of Inertia
Now, we add the moments of inertia of both halves:
\[ I_{total} = I_1 + I_2 = \frac{ML^2}{24} + \frac{ML^2}{6} \]
To combine these, we need a common denominator:
\[ I_{total} = \frac{ML^2}{24} + \frac{4ML^2}{24} = \frac{5ML^2}{24} \]

### Final Answer
Thus, the moment of inertia of the bent rod about the specified axis is:
\[ \boxed{\frac{5ML^2}{24}} \]