At certain temperature, $\operatorname{rod} \mathrm{A}$ and $\operatorname{rod} \mathrm{B}$ of different…
At certain temperature, $\operatorname{rod} \mathrm{A}$ and $\operatorname{rod} \mathrm{B}$ of different materials have lengths $L_A$ and $L_B$ respectively: Their coefficients of linear expansion are $\alpha_A$ and $\alpha_B$ respectively. It is observed that the difference between their length's remains constant at all temperatures. The ratio $\mathrm{L}_{\mathrm{A}}: \mathrm{L}_{\mathrm{B}}$ is given by
$\frac{\alpha_A}{\alpha_B}$
$\frac{\alpha_B}{\alpha_A}$
$\frac{\alpha_A+\alpha_B}{\alpha_A}$
$\frac{\alpha_A+\alpha_B}{\alpha_B}$
Solution
Length of $\operatorname{rod} \mathrm{A}$ at temperature $\mathrm{t}=\mathrm{L}_{\mathrm{A}}+l_{\mathrm{A}} \alpha_{\mathrm{A}} \Delta \mathrm{t}$
Length of $\operatorname{rod} \mathrm{B}$ at temperature $\mathrm{t}=\mathrm{L}_{\mathrm{B}}+l_{\mathrm{B}} \alpha_{\mathrm{B}} \Delta \mathrm{t}$
Length of $\operatorname{rod} A-$ length of $\operatorname{rod} B$
$=\left(\mathrm{L}_{\mathrm{A}}-\mathrm{L}_{\mathrm{B}}\right)+\left(l_{\mathrm{A}} \alpha_{\mathrm{A}}-l_{\mathrm{B}} \alpha_{\mathrm{B}}\right) \Delta \mathrm{t}$ For difference in the length to be constant, coefficient of $\Delta t$ must be zero.
$\begin{array}{ll}
\therefore & \quad l_{\mathrm{A}} \alpha_{\mathrm{A}}-l_{\mathrm{B}} \alpha_{\mathrm{B}}=0 \\
\therefore & l_{\mathrm{A}} \alpha_{\mathrm{A}}=l_{\mathrm{B}} \alpha_{\mathrm{B}} \\
\therefore & \frac{l_{\mathrm{A}}}{l_{\mathrm{B}}}=\frac{\alpha_{\mathrm{B}}}{\alpha_{\mathrm{A}}}
\end{array}$