A wire is made by attaching two segments together end to end. One segment is made of aluminium and the other is steel. The effective coefficient of linear expansion of the two segment wire is `19xx10^(-6)//^(@)C`. The fraction length of aluminium is (linear coefficient of thermal expansion of aluminium and steel are `23xx10^(-6).^(@)C` and `12xx10^(-6)//^(@)C`,

Answer

To solve the problem, we need to find the fraction length of the aluminum segment in a wire made of two segments: aluminum and steel. We are given the coefficients of linear expansion for both materials and the effective coefficient of linear expansion for the combined wire.

### Step-by-Step Solution:

1. **Identify Given Values:**
- Coefficient of linear expansion of aluminum, \( \alpha_{Al} = 23 \times 10^{-6} \, ^{\circ}C^{-1} \)
- Coefficient of linear expansion of steel, \( \alpha_{Steel} = 12 \times 10^{-6} \, ^{\circ}C^{-1} \)
- Effective coefficient of linear expansion of the combined wire, \( \alpha_{eff} = 19 \times 10^{-6} \, ^{\circ}C^{-1} \)

2. **Set Up the Equation for Effective Coefficient:**
The effective coefficient of linear expansion for two segments can be expressed as:
\[ \alpha_{eff} = \frac{L_1 \alpha_{Al} + L_2 \alpha_{Steel}}{L_1 + L_2} \]
where \( L_1 \) is the length of the aluminum segment and \( L_2 \) is the length of the steel segment.

3. **Express Lengths in Terms of Total Length:**
Let \( L = L_1 + L_2 \). We can express \( L_1 \) and \( L_2 \) as:
\[ L_1 = xL \quad \text{and} \quad L_2 = (1-x)L \]
where \( x \) is the fraction of the total length that is aluminum.

4. **Substitute into the Effective Coefficient Equation:**
Substitute \( L_1 \) and \( L_2 \) into the equation:
\[ \alpha_{eff} = \frac{xL \alpha_{Al} + (1-x)L \alpha_{Steel}}{L} \]
Simplifying this gives:
\[ \alpha_{eff} = x \alpha_{Al} + (1-x) \alpha_{Steel} \]

5. **Rearranging the Equation:**
Rearranging the equation, we have:
\[ \alpha_{eff} = x \alpha_{Al} + \alpha_{Steel} - x \alpha_{Steel} \]
\[ \alpha_{eff} = x (\alpha_{Al} - \alpha_{Steel}) + \alpha_{Steel} \]
\[ x (\alpha_{Al} - \alpha_{Steel}) = \alpha_{eff} - \alpha_{Steel} \]
\[ x = \frac{\alpha_{eff} - \alpha_{Steel}}{\alpha_{Al} - \alpha_{Steel}} \]

6. **Substituting the Values:**
Now substitute the known values:
\[ x = \frac{19 \times 10^{-6} - 12 \times 10^{-6}}{23 \times 10^{-6} - 12 \times 10^{-6}} \]
\[ x = \frac{7 \times 10^{-6}}{11 \times 10^{-6}} = \frac{7}{11} \]

7. **Final Result:**
Therefore, the fraction length of aluminum in the wire is:
\[ x = \frac{7}{11} \]