The coefficient of linear expansion of brass and steel rods are $\alpha_1$ and $\alpha_2$ respectively.…
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?
$\alpha_1 l_2=\alpha_2 l_1$
$\alpha_1^2 l_2=\alpha_2^2 l_1$
$\alpha_{1} l_2^2=\alpha_2 l_1^2$
$l_1 \alpha_1=l_2 \alpha_2$
Solution
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.