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 find the moment of inertia of the cross formed by two rods about an axis passing through their common center and perpendicular to the plane formed by them, we can follow these steps:

### Step 1: Understand the Configuration
We have two rods, each of mass \( m \) and length \( L = 1 \), joined at their centers to form a cross. The rods are positioned at right angles to each other.

### Step 2: Identify the Axis of Rotation
The axis of rotation is perpendicular to the plane formed by the two rods and passes through their common center.

### Step 3: Use the Perpendicular Axis Theorem
The perpendicular axis theorem states that for a planar body, the moment of inertia about an axis perpendicular to the plane (Iz) is equal to the sum of the moments of inertia about two perpendicular axes in the plane (Ix and Iy):
\[ I_z = I_x + I_y \]

### Step 4: Calculate the Moment of Inertia for Each Rod
1. **For the rod along the x-axis:**
The moment of inertia about its own axis (which is along its length) is:
\[ I_{x} = 0 \quad \text{(since the axis is along the length)} \]
The moment of inertia about an axis perpendicular to its length and passing through its center is given by:
\[ I_{x, \text{perpendicular}} = \frac{mL^2}{12} \]
Since the rod is along the x-axis, we consider the moment of inertia about the y-axis:
\[ I_{y} = \frac{mL^2}{12} \]

2. **For the rod along the y-axis:**
Similarly, for the rod along the y-axis:
\[ I_{y} = 0 \quad \text{(since the axis is along the length)} \]
The moment of inertia about an axis perpendicular to its length and passing through its center is:
\[ I_{y, \text{perpendicular}} = \frac{mL^2}{12} \]
Since the rod is along the y-axis, we consider the moment of inertia about the x-axis:
\[ I_{x} = \frac{mL^2}{12} \]

### Step 5: Combine the Moments of Inertia
Using the perpendicular axis theorem:
\[ I_z = I_x + I_y = \left(\frac{mL^2}{12}\right) + \left(\frac{mL^2}{12}\right) = \frac{2mL^2}{12} = \frac{mL^2}{6} \]

### Step 6: Substitute the Length
Since the length \( L = 1 \):
\[ I_z = \frac{m(1)^2}{6} = \frac{m}{6} \]

### Final Answer
The moment of inertia of the cross about the axis passing through the common center of the rods and perpendicular to the plane formed by them is:
\[ \boxed{\frac{m}{6}} \]