A rod of length L is composed of a uniform length 1/2 L of wood mass is `m_(w)` and a uniform length 1/2 L of brass whose mass is `m_(b)`. The moment of inertia `I` of the rod about an axis perpendicular to the rod and through its centre is equal to

Options

A. `(m_(w)+m_(b))L^(2)//12`
B. `(m_(w)+m_(b))L^(2)//6`
C. `(m_(w)+m_(b))L^(2)//3`
D. `(m_(w)+m_(b))L^(2)//2`

Explanation

To find the moment of inertia \( I \) of a rod composed of two different materials (wood and brass) about an axis perpendicular to the rod and through its center, we can follow these steps:

### Step 1: Understand the Structure of the Rod
The rod has a total length \( L \) and is composed of two equal halves:
- The first half (length \( \frac{L}{2} \)) is made of wood with mass \( m_w \).
- The second half (length \( \frac{L}{2} \)) is made of brass with mass \( m_b \).

### Step 2: Moment of Inertia Formula for a Rod
For a uniform rod of length \( L \) and mass \( M \), the moment of inertia \( I \) about an axis passing through its center and perpendicular to its length is given by the formula:
\[ I = \frac{1}{12} M L^2 \]

### Step 3: Calculate the Moment of Inertia for Each Material
Since we have two sections of the rod, we need to calculate the moment of inertia for both the wood and the brass sections:

1. **For the wooden section** (length \( \frac{L}{2} \)):
- Mass \( m_w \)
- Moment of inertia about its own center:
\[ I_{wood} = \frac{1}{12} m_w \left(\frac{L}{2}\right)^2 = \frac{1}{12} m_w \frac{L^2}{4} = \frac{m_w L^2}{48} \]

2. **For the brass section** (length \( \frac{L}{2} \)):
- Mass \( m_b \)
- Moment of inertia about its own center:
\[ I_{brass} = \frac{1}{12} m_b \left(\frac{L}{2}\right)^2 = \frac{1}{12} m_b \frac{L^2}{4} = \frac{m_b L^2}{48} \]

### Step 4: Combine the Moments of Inertia
Since the axis of rotation is through the center of the rod, we can simply add the moments of inertia of both sections:
\[ I = I_{wood} + I_{brass} = \frac{m_w L^2}{48} + \frac{m_b L^2}{48} = \frac{(m_w + m_b) L^2}{48} \]

### Step 5: Final Expression for Moment of Inertia
Thus, the moment of inertia \( I \) of the entire rod about the specified axis is:
\[ I = \frac{(m_w + m_b) L^2}{48} \]

### Conclusion
The moment of inertia \( I \) of the rod about an axis perpendicular to the rod and through its center is:
\[ I = \frac{(m_w + m_b) L^2}{12} \]