Two rods each of mass m and length 1 are joined at the centre to form a cross. The moment of inertia of this cross about an axis passing through the common centre of the rods and perpendicular to the plane formed by them, is :

Options

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

Answer

To solve the problem of finding the moment of inertia of a cross formed by two rods, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Each rod has a mass \( m \).
- Each rod has a length \( l \) (we will denote this as \( l \) instead of 1 for generality).

2. **Understand the Configuration:**
- The two rods are joined at their centers, forming a cross shape.
- We need to find the moment of inertia about an axis that passes through the common center of the rods and is perpendicular to the plane formed by them.

3. **Moment of Inertia of a Single Rod:**
- The moment of inertia \( I \) of a single rod about an axis through its center and perpendicular to its length is given by the formula:
\[ I = \frac{1}{12} m l^2 \]
- This formula accounts for the distribution of mass along the length of the rod.

4. **Calculate the Moment of Inertia for Both Rods:**
- Since both rods are identical, the moment of inertia for each rod about the same axis is:
\[ I_1 = \frac{1}{12} m l^2 \]
\[ I_2 = \frac{1}{12} m l^2 \]

5. **Total Moment of Inertia:**
- The total moment of inertia \( I_{total} \) for the cross configuration is the sum of the moments of inertia of both rods:
\[ I_{total} = I_1 + I_2 = \frac{1}{12} m l^2 + \frac{1}{12} m l^2 \]
- Simplifying this gives:
\[ I_{total} = \frac{2}{12} m l^2 = \frac{1}{6} m l^2 \]

6. **Final Result:**
- Therefore, the moment of inertia of the cross about the specified axis is:
\[ I_{total} = \frac{1}{6} m l^2 \]

### Conclusion:
The moment of inertia of the cross formed by the two rods about the given axis is \( \frac{1}{6} m l^2 \).