Coefficient of linear expansion of brass and steel rods are α_1 and...

Let the initial lengths of the brass and steel rods be $l_1$ and $l_2$ respectively at temperature $T$.

When the temperature is increased by $\Delta T$, the new lengths are:
$l_1' = l_1(1 + \alpha_1 \Delta T)$
$l_2' = l_2(1 + \alpha_2 \Delta T)$

The difference in lengths at the new temperature is:
$l_2' - l_1' = l_2(1 + \alpha_2 \Delta T) - l_1(1 + \alpha_1 \Delta T)$
$l_2' - l_1' = (l_2 - l_1) + (l_2 \alpha_2 - l_1 \alpha_1) \Delta T$

Given that the difference in lengths $(l_2 - l_1)$ is maintained the same at all temperatures, we have:
$l_2' - l_1' = l_2 - l_1$

Substituting this into the equation gives:
$(l_2 \alpha_2 - l_1 \alpha_1) \Delta T = 0$

Since $\Delta T \neq 0$, we get:
$l_1 \alpha_1 = l_2 \alpha_2$

Answer: $\alpha_1 l_1 = \alpha_2 l_2$