A wire is made by attaching two segments together end to end. One segment is made of aluminum and the other is steel. The effective coefficient of linear expansion of the two segment wire is $19\times {10}^{-6}/\left({ }^{\circ }\mathrm{C}\right)$. The fraction length of aluminum is [linear coefficients of thermal expansion of aluminum and steel are $23\times {10}^{-6}/\left({ }^{\circ }\mathrm{C}\right)$ and $12\times {10}^{-6}/\left({ }^{\circ }\mathrm{C}\right)$,respectively]

The change in length of wire : $={\mathrm{l}}_{\mathrm{Al}}{\mathrm{\alpha }}_{\mathrm{Al}}\mathrm{\Delta \theta }+{\mathrm{l}}_{\mathrm{st}}{\mathrm{\alpha }}_{\mathrm{st}}\mathrm{\Delta \theta }$ associated with temperature change of $\mathrm{\Delta \theta }$, where ${\mathrm{\alpha }}_{\mathrm{Al}}$ and ${\mathrm{\alpha }}_{\mathrm{St}}$ are the coefficient of linear expansion of aluminum and steel, respectively 

$\begin{array}{l}{\mathrm{\alpha }}_{\mathrm{Al}}=23\times {10}^{-6}{/}^{\circ }\mathrm{C}\\ {\mathrm{\alpha }}_{\mathrm{st}}=12\times {10}^{-6}{/}^{\circ }\mathrm{C}\end{array}$

The effective coefficient of linear expansion of the two segments of wire $=19\times {10}^{-6}{/}^{\circ }\mathrm{C}$

$\begin{array}{l}{\mathrm{l}}_1{\mathrm{\alpha }}_{\mathrm{Al}}\mathrm{\Delta \theta }+{\mathrm{l}}_2{\mathrm{\alpha }}_{\mathrm{st}}\mathrm{\Delta \theta }=\left({\mathrm{l}}_1+{\mathrm{l}}_2\right)\mathrm{\alpha \Delta \theta }\\ \frac{{\mathrm{l}}_1}{\left({\mathrm{l}}_1+{\mathrm{l}}_2\right)}=\frac{\mathrm{\alpha }-\frac{{\mathrm{l}}_2}{\left({\mathrm{l}}_1+{\mathrm{l}}_2\right)}{\mathrm{\alpha }}_{\mathrm{st}}}{{\mathrm{\alpha }}_{\mathrm{Al}}}\\ \left[\frac{{\mathrm{l}}_1}{{\mathrm{l}}_1+{\mathrm{l}}_2}=\mathrm{x},{\mathrm{l}}_1+{\mathrm{l}}_2=\mathrm{l},\left(\frac{{\mathrm{l}}_2}{{\mathrm{l}}_1+{\mathrm{l}}_2}\right)=1-\mathrm{x}\right]\end{array}$

$\begin{array}{l}\mathrm{x}=\frac{\mathrm{\alpha }-(1-\mathrm{x}){\mathrm{\alpha }}_{\mathrm{st}}}{{\mathrm{\alpha }}_{\mathrm{Al}}}\\ \mathrm{x}=\frac{\mathrm{\alpha }-{\mathrm{\alpha }}_{\mathrm{st}}}{{\mathrm{\alpha }}_{\mathrm{Al}}-{\mathrm{\alpha }}_{\mathrm{st}}}=\frac{19\times {10}^{-6}-12\times {10}^{-6}}{23\times {10}^{-6}-12\times {10}^{-6}}=\frac{7}{11}\end{array}$