The coefficient of linear expansion of brass and steel are α₁ and α₂. If we take a brass rod of length L₁ and steel rod of length L₂ at 0°C , their difference in length ( L₂-L₁ ) will remain the same at any temperature if

The correct option is α₁L₁ = α₂L₂.

To understand why this is the correct condition, let us look at how the lengths of the brass and steel rods change with temperature.

The length of a rod at any temperature change $\Delta T$ is given by the formula for linear thermal expansion:
$L_T=L_0(1+\alpha \Delta T)$
where $L_0$ is the initial length at 0°C, $\alpha$ is the coefficient of linear expansion, and $L_T$ is the length at temperature T.

For the brass rod of initial length $L_1$ and coefficient of linear expansion ${\alpha }_1$, the length at temperature T is:
$L_{1T}=L_1(1+{\alpha }_1\Delta T)=L_1+L_1{\alpha }_1\Delta T$

Similarly, for the steel rod of initial length $L_2$ and coefficient of linear expansion ${\alpha }_2$, the length at temperature T is:
$L_{2T}=L_2(1+{\alpha }_2\Delta T)=L_2+L_2{\alpha }_2\Delta T$

The difference between the two lengths at temperature T is:
$L_{2T}-L_{1T}=(L_2+L_2{\alpha }_2\Delta T)-(L_1+L_1{\alpha }_1\Delta T)$

Rearranging the terms, we get:
$L_{2T}-L_{1T}=(L_2-L_1)+(L_2{\alpha }_2-L_1{\alpha }_1)\Delta T$

For the difference in length to remain constant at all temperatures, the difference $L_{2T}-L_{1T}$ must be equal to the initial difference $L_2-L_1$ regardless of the value of $\Delta T$. This is only possible if the temperature-dependent term is zero:
$(L_2{\alpha }_2-L_1{\alpha }_1)=0$

This gives:
${\alpha }_1{L}_1={\alpha }_2{L}_2$

Therefore, the difference in length remains the same at any temperature if ${\alpha }_1{L}_1={\alpha }_2{L}_2$.