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 two rods of equal lengths \( L \) and equal masses \( M \) placed along the x-axis and y-axis respectively, about the line \( y = x \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Configuration**:
- We have two rods: one along the x-axis and the other along the y-axis.
- Both rods have mass \( M \) and length \( L \).
- The center of mass of the system is at the origin (0,0).

2. **Moment of Inertia about the Axes**:
- The moment of inertia \( I_x \) of a rod about an axis perpendicular to it through its center is given by:
\[ I_x = \frac{1}{12} M L^2 \]
- For the rod along the x-axis, the moment of inertia about the x-axis is:
\[ I_{x} = \frac{1}{12} M L^2 \]
- For the rod along the y-axis, the moment of inertia about the y-axis is:
\[ I_{y} = \frac{1}{12} M L^2 \]

3. **Using the Perpendicular Axis Theorem**:
- The Perpendicular Axis Theorem states that for a planar body:
\[ I_z = I_x + I_y \]
- Here, \( I_z \) is the moment of inertia about an axis perpendicular to the plane of the rods (z-axis).
- Therefore, we have:
\[ I_z = \frac{1}{12} M L^2 + \frac{1}{12} M L^2 = \frac{1}{6} M L^2 \]

4. **Finding Moment of Inertia about the Line \( y = x \)**:
- The line \( y = x \) can be treated as a new axis. To find the moment of inertia about this line, we can use the fact that the moment of inertia about any axis can be calculated using the relationship:
\[ I_{y=x} = \frac{1}{2}(I_x + I_y) \]
- Since \( I_x = I_y \) for our configuration, we can write:
\[ I_{y=x} = \frac{1}{2} \left( \frac{1}{12} M L^2 + \frac{1}{12} M L^2 \right) = \frac{1}{2} \left( \frac{1}{6} M L^2 \right) = \frac{1}{12} M L^2 \]

5. **Conclusion**:
- The moment of inertia of the system about the line \( y = x \) is:
\[ I_{y=x} = \frac{1}{12} M L^2 \]

### Final Answer:
The moment of inertia about the line \( y = x \) is \( \frac{1}{12} M L^2 \).

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