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`,

Options

A. `(5)/(11)`
B. `(6)/(11)`
C. `(7)/(11)`
D. `(8)/(11)`

Explanation

To solve the problem, we need to find the fraction of the length of aluminum in a wire made of two segments: aluminum and steel. We know the effective coefficient of linear expansion for the combined wire and the coefficients for each material.

### Step-by-Step Solution:

1. **Define Variables:**
- Let \( L_1 \) be the length of the aluminum segment.
- Let \( L_2 \) be the length of the steel segment.
- The coefficient of linear expansion for aluminum, \( \alpha_1 = 23 \times 10^{-6} \, ^{\circ}C^{-1} \).
- The coefficient of linear expansion for steel, \( \alpha_2 = 12 \times 10^{-6} \, ^{\circ}C^{-1} \).
- The effective coefficient of linear expansion for the combined wire, \( \alpha_{\text{effective}} = 19 \times 10^{-6} \, ^{\circ}C^{-1} \).

2. **Write the Equation for Effective Length Change:**
The total effective length change can be expressed as:
\[ \Delta L_{\text{effective}} = (L_1 + L_2) \alpha_{\text{effective}} \Delta T \]
This can also be expressed as the sum of the individual length changes:
\[ \Delta L_{\text{effective}} = L_1 \alpha_1 \Delta T + L_2 \alpha_2 \Delta T \]

3. **Cancel \(\Delta T\):**
Since \(\Delta T\) appears in all terms, we can cancel it out:
\[ (L_1 + L_2) \alpha_{\text{effective}} = L_1 \alpha_1 + L_2 \alpha_2 \]

4. **Substitute Known Values:**
Substitute the values of \(\alpha_{\text{effective}}\), \(\alpha_1\), and \(\alpha_2\):
\[ (L_1 + L_2) (19 \times 10^{-6}) = L_1 (23 \times 10^{-6}) + L_2 (12 \times 10^{-6}) \]

5. **Eliminate the Common Factor:**
Divide through by \(10^{-6}\):
\[ (L_1 + L_2) \cdot 19 = L_1 \cdot 23 + L_2 \cdot 12 \]

6. **Rearranging the Equation:**
Rearranging gives:
\[ 19L_1 + 19L_2 = 23L_1 + 12L_2 \]
Simplifying this leads to:
\[ 19L_1 + 19L_2 - 23L_1 - 12L_2 = 0 \]
Which simplifies to:
\[ -4L_1 + 7L_2 = 0 \]

7. **Express \(L_2\) in Terms of \(L_1\):**
From the above equation, we can express \(L_2\) in terms of \(L_1\):
\[ 4L_1 = 7L_2 \implies L_2 = \frac{4}{7}L_1 \]

8. **Finding the Fraction of Aluminum:**
The total length \(L = L_1 + L_2\) can be expressed as:
\[ L = L_1 + \frac{4}{7}L_1 = \frac{11}{7}L_1 \]
Therefore, the fraction of the length of aluminum is:
\[ \frac{L_1}{L} = \frac{L_1}{\frac{11}{7}L_1} = \frac{7}{11} \]

### Final Answer:
The fraction length of aluminum in the wire is \( \frac{7}{11} \).

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