The coefficient of linear expansion of brass and steel rods are \(\alpha_1\) and \(\alpha_2\) respectively. Lengths of brass and steel rods are \(l_1\) and \(l_2\) respectively. If \(\left(l_2-l_1\right)\) is maintained same at all temperatures, which one of the following relations is correct?

The correct option is (D).
Concept: Length of a metalic rod at a given temperature is given by,
\(l=l_0(1+\alpha \Delta T)\)
where, \(l_0\) is the initial length, \(\alpha\) the coefficient of linear expansion and \(\Delta T\) the temperature change.
At any given temperature the lengths of the brass and steel rods can be written as, \(l_b=l_1\left(1+\alpha_1 \Delta T\right)\) and \(l_s=l_2\left(1+\alpha_2 \Delta T\right)\) respectively.
Therefore, at difference of length is given by,
\(\begin{aligned} & l_b-l_s=l_1\left(1+\alpha_1 \Delta T\right)-l_2\left(1+\alpha_2 \Delta T\right) \\ & \Rightarrow l_b-l_s=\left(l_1-l_2\right)+\left(\alpha_1 l_1-l_2 \alpha_2\right) \Delta T\end{aligned}\)
Since, \(l_b-l_{\mathrm{s}}=l_1-l_2\) as the difference of length remains constant.
Therefore, \(\left(\alpha_1 l_1-l_2 \alpha_2\right)=0\)
So, \(\alpha_1 l_1=l_2 \alpha_2\) is the answer.
Since, \(l_b-l_s=l_1-l_2\) as the difference of length remains constant.
Therefore, \(\left(\alpha_1 l_1-l_2 \alpha_2\right)=0\) So, \(\alpha_1 l_1=l_2 \alpha_2\) is the answer.