Two rods of equal lengths(l) and equal mass M are kept along x and y axis respectively such that their centre of mass lie at origin. The moment of inertia about an line y = x, is

Options

A. `(ml^(2))/(3)`
B. `(ml^(2))/(4)`
C. `(ml^(2))/(12)`
D. `(ml^(2))/(6)`

Answer

To find the moment of inertia of the two rods about the line \( y = x \), we can follow these steps:

### Step 1: Understand the Configuration
We have two rods of equal length \( l \) and equal mass \( M \). One rod is aligned along the x-axis and the other along the y-axis. Their centers of mass are at the origin (0,0).

### Step 2: Identify the Angle
The line \( y = x \) makes an angle of \( 45^\circ \) with both the x-axis and y-axis.

### Step 3: Moment of Inertia Formula
The moment of inertia \( I \) of a rod about an axis passing through its center of mass and making an angle \( \theta \) with the rod is given by:
\[ I = \frac{ML^2}{12} \sin^2 \theta \]
where \( L \) is the length of the rod.

### Step 4: Calculate Moment of Inertia for Each Rod
1. **For the rod along the x-axis:**
- The angle \( \theta = 45^\circ \).
- Thus, the moment of inertia \( I_1 \) is:
\[ I_1 = \frac{ML^2}{12} \sin^2(45^\circ) = \frac{ML^2}{12} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{ML^2}{12} \cdot \frac{1}{2} = \frac{ML^2}{24} \]

2. **For the rod along the y-axis:**
- The angle \( \theta = 45^\circ \) is the same.
- Thus, the moment of inertia \( I_2 \) is:
\[ I_2 = \frac{ML^2}{12} \sin^2(45^\circ) = \frac{ML^2}{12} \cdot \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{ML^2}{12} \cdot \frac{1}{2} = \frac{ML^2}{24} \]

### Step 5: Total Moment of Inertia
Since the moment of inertia is a scalar quantity, we can add the moments of inertia of both rods:
\[ I_{total} = I_1 + I_2 = \frac{ML^2}{24} + \frac{ML^2}{24} = \frac{2ML^2}{24} = \frac{ML^2}{12} \]

### Final Answer
The moment of inertia of the system about the line \( y = x \) is:
\[ \boxed{\frac{ML^2}{12}} \]