Coefficient of linear expansion of brass and steel rods are  ${\alpha }_1$and ${\alpha }_2$ Lengths of brass and steel rods are ${\mathcal{l}}_1$and ${\mathcal{l}}_2$respectively. If $({\mathcal{l}}_2-{\mathcal{l}}_1)$ is maintained same at all temperatures, which one of the following relations holds good?

$\text{ Linear expansion of brass }={\alpha }_1$

$\text{ Linear expansion of steel }={\alpha }_2$

$\text{ Length of brass rod }={\mathcal{l}}_1$

$\text{ Length of steel rod }={\mathcal{l}}_2$

$On increa\sin g the temperature of the rods by ∆T, new lengths would be$

${\mathcal{l}}_1^{\text{'}}={\mathcal{l}}_1\left(1+{\alpha }_1\Delta T\right)........\left(1\right)$

${\mathcal{l}}_2^{\text{'}}={\mathcal{l}}_2\left(1+{\alpha }_2\Delta T\right).........\left(2\right)$

$\text{ Subtracting eqn. (i) from eqn. (ii), we get }$

${\mathcal{l}}_2^{\text{'}}-{\mathcal{l}}_1^{\text{'}}=\left({\mathcal{l}}_2-{\mathcal{l}}_1\right)+\left({\mathcal{l}}_2{\alpha }_2-{\mathcal{l}}_1{\alpha }_1\right)\Delta T$

According to the question

$\begin{array}{c} {\mathcal{l}}_2^{\text{'}}-{\mathcal{l}}_1^{\text{'}}={\mathcal{l}}_2-{\mathcal{l}}_1\\ \therefore {\mathcal{l}}_2{\alpha }_2-{\mathcal{l}}_1{\alpha }_1=0\text{ or }{\mathcal{\text{l}}}_1{\alpha }_1={\mathcal{l}}_2{\alpha }_2\end{array}$