Question

Two rods of length ${l_1}$ and ${l_2}$ are made of materials whose coefficients of linear expansion are ${\alpha _1}$ and ${\alpha _2}$. If the difference between the two lengths is independent of temperature, then:
A. $\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{\alpha _1}}}{{{\alpha _2}}}$
B. $\dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}}$
C. $l_2^2{\alpha _1} = l_1^2{\alpha _2}$
D. $\dfrac{{\alpha _1^2}}{{{l_1}}} = \dfrac{{\alpha _2^2}}{{{l_2}}}$

Answer 1

Since the change in length of the rods is independent of temperature, the change in length for the both rods will be the same even if they have different coefficient of expansions. Express the change in length of the rods using the expression for linear expansion of the material with respect to temperature. Equate the change in lengths for the two rods.

Formula used:
$\Delta l = {l_1}\alpha \Delta T$
Here, ${l_1}$ is the original length of the rod, $\Delta l$ is the change in length of the rod, $\alpha$ is the coefficient of linear expansion for the rod and $\Delta T$ is the change in temperature.

Complete step by step answer:
We have given that the change in length of the rods is independent of temperature. Therefore, the change in length for the both rods will be the same even if they have different coefficients of expansions. Let us express the equation for the linear expansion of the rod of length ${l_1}$ as follows,
$\Delta {l_1} = {l_1}{\alpha _1}\Delta T$ ……. (1)
Here, ${l_1}$ is the original length of the first rod, $\Delta {l_1}$ is the change in length of the first rod, ${\alpha _1}$ is the coefficient of linear expansion for the first rod and $\Delta T$ is the change in temperature.
Also, for the second rod,
$\Delta {l_2} = {l_2}{\alpha _2}\Delta T$ ……. (2)
Here, ${l_2}$ is the original length of the second rod, $\Delta {l_2}$ is the change in length of the second rod, ${\alpha _2}$ is the linear expansion coefficient of the second rod and $\Delta T$ is the change in temperature.
According to the question, we have,
$\Delta {l_1} = \Delta {l_2}$
Using equation (1) and (2) in the above equation, we get,
${l_1}{\alpha _1}\Delta T = {l_2}{\alpha _2}\Delta T$
Since the temperature change is the same for the rods, we get,
${l_1}{\alpha _1} = {l_2}{\alpha _2}$
$\therefore \dfrac{{{l_1}}}{{{l_2}}} = \dfrac{{{\alpha _2}}}{{{\alpha _1}}}$

So, the correct answer is option A.

Note: The linear expansion of the material is actually expressed as ${l_2} = {l_1} + {l_1}\alpha \left( {{T_2} - {T_1}} \right)$, where, ${l_2}$ is the new length, ${l_1}$ is the original length, $\alpha$ is the coefficient of linear expansion, $T$ is the initial temperature and ${T_2}$ is the final temperature. If the final temperature is greater than the initial temperature, the material tends to increase in its length and if the final temperature is less than the initial temperature, the material undergoes compression.