The coefficient of linear expansion of an in homogeneous rod change linearly from `alpha_(1)` to `alpha_(2)` from one end to the other end of the rod. The effective coefficient of linear expansion of rod is

Options

A. `alpha_1+alpha_2`
B. `(alpha_1+alpha_2)/(2)`
C. `sqrt(alpha_1alpha_2)`
D. `alpha_1-alpha_2`

Explanation

To find the effective coefficient of linear expansion of an inhomogeneous rod whose coefficient of linear expansion changes linearly from \( \alpha_1 \) to \( \alpha_2 \) from one end to the other, we can follow these steps:

### Step-by-Step Solution:

1. **Understanding the Problem**:
The coefficient of linear expansion \( \alpha(x) \) at a distance \( x \) from one end of the rod is given by:
\[ \alpha(x) = \alpha_1 + \left( \frac{\alpha_2 - \alpha_1}{L} \right) x \]
where \( L \) is the length of the rod.

2. **Expressing Change in Length**:
The change in length \( \Delta L \) of a small segment \( dx \) of the rod due to a temperature change \( \Delta T \) can be expressed as:
\[ dL = \alpha(x) \cdot dx \cdot \Delta T \]

3. **Integrating Over the Length of the Rod**:
To find the total change in length \( \Delta L \) of the entire rod, we integrate \( dL \) from \( 0 \) to \( L \):
\[ \Delta L = \int_0^L \alpha(x) \cdot dx \cdot \Delta T \]

4. **Substituting \( \alpha(x) \)**:
Substitute \( \alpha(x) \) into the integral:
\[ \Delta L = \Delta T \int_0^L \left( \alpha_1 + \left( \frac{\alpha_2 - \alpha_1}{L} \right) x \right) dx \]

5. **Calculating the Integral**:
Break the integral into two parts:
\[ \Delta L = \Delta T \left( \int_0^L \alpha_1 \, dx + \int_0^L \left( \frac{\alpha_2 - \alpha_1}{L} \right) x \, dx \right) \]
The first integral evaluates to:
\[ \int_0^L \alpha_1 \, dx = \alpha_1 L \]
The second integral evaluates to:
\[ \int_0^L \left( \frac{\alpha_2 - \alpha_1}{L} \right) x \, dx = \frac{\alpha_2 - \alpha_1}{L} \cdot \frac{L^2}{2} = \frac{(\alpha_2 - \alpha_1)L}{2} \]

6. **Combining the Results**:
Combine the results of the integrals:
\[ \Delta L = \Delta T \left( \alpha_1 L + \frac{(\alpha_2 - \alpha_1)L}{2} \right) \]
Simplifying gives:
\[ \Delta L = \Delta T \left( \frac{2\alpha_1 + \alpha_2 - \alpha_1}{2} L \right) = \Delta T \left( \frac{\alpha_1 + \alpha_2}{2} L \right) \]

7. **Finding the Effective Coefficient of Linear Expansion**:
The effective coefficient of linear expansion \( \alpha \) can be defined as:
\[ \Delta L = \alpha \cdot L \cdot \Delta T \]
By equating both expressions for \( \Delta L \):
\[ \alpha \cdot L \cdot \Delta T = \Delta T \left( \frac{\alpha_1 + \alpha_2}{2} L \right) \]
Dividing both sides by \( L \cdot \Delta T \) (assuming \( L \) and \( \Delta T \) are not zero):
\[ \alpha = \frac{\alpha_1 + \alpha_2}{2} \]

### Final Answer:
The effective coefficient of linear expansion of the rod is:
\[ \alpha = \frac{\alpha_1 + \alpha_2}{2} \]