At certain temperature, rod A and $\operatorname{rod} \mathrm{B}$ of different materials have lengths $L_A$…
At certain temperature, rod A and $\operatorname{rod} \mathrm{B}$ of different materials have lengths $L_A$ and $L_B$ respectively. Their co-efficients of linear expansion are $\alpha_A$ and $\alpha_B$ respectively. It is observed that the difference between their lengths remain constant at all temperatures. The ratio $L_A / L_B$ 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
\(L_B^{\prime}=L_B\left(1+\alpha_B \Delta \theta\right)\) and \(L_A^{\prime}=L_A\left(1+\alpha_B \Delta \theta\right)\). So that
\(L_B^{\prime}-L_A^{\prime}=L_B-L_S+\left(L_B \alpha_B-L_A \alpha_A\right) \Delta \theta\)
So \(\left(L_B^{\prime}-L_A^{\prime}\right)\) will be equal to \(\left(L_B-L_A\right)\) at all temperature if, \(L_B \alpha_B-L_A \alpha_A=0\)
\([\operatorname{as} \Delta \theta \neq 0] \text { or } \frac{L_B}{L_A}=\frac{\alpha_{A}}{\alpha_B}\)
i.e., the difference in the lengths of the two rods will be independent of temperature if the lengths are in the inverse ratio of their coefficients of linear expansion.