Question

Two uniform metal rods of length ${{l}_{1}}$ and ${{l}_{2}}$ and the linear coefficients of expansion ${{\alpha }_{1}}$ and ${{\alpha }_{2}}$ respectively are connected to form a single rod of length ${{l}_{1}}+{{l}_{2}}$. When the temperature of the combined rod is raised by $t{}^\circ \text{C}$, the length of each rod increases by the same amount. Then $\dfrac{{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}$ is:
A. $\dfrac{{{l}_{1}}}{{{l}_{1}}+{{l}_{2}}}$
B. $\dfrac{{{l}_{1}}+{{l}_{2}}}{{{l}_{1}}}$
C. $\dfrac{{{l}_{2}}}{{{l}_{1}}+{{l}_{2}}}$
D. $\dfrac{{{l}_{1}}+{{l}_{2}}}{{{l}_{2}}}$

Answer 1

Hint: We can calculate this problem by using the coefficient of linear expansion formula. By comparing the two metals expansion behaviour, we can find the change in the length when they combined. Since the temperature is common to both materials, we don’t have to consider that in the problem.

Formula used:
$\Delta l=\alpha L\Delta T$, where $\Delta l$ is the change in length of the material, L is the original length of the material, $\alpha$ is the linear coefficient of expansion and $\Delta T$ is the change in temperature.

Complete step by step answer:
Linear expansion is the expansion of the length of the material. Here, the expansion is happening due to the change in temperature. So, the change in temperature will make an impact on the rate of expansion.
According to the linear coefficient of expansion of metals, the length is inversely proportional to the coefficient of linear expansion.
Here we are combining the two rods and applying the same temperature. We can observe both rods are expanding by the same amount of length. So, we can write the new length as of first road as,
${{L}_{1}}={{l}_{1}}+{{l}_{1}}{{\alpha }_{1}}\Delta T$
The new length of the second road will be,
${{L}_{2}}={{l}_{2}}+{{l}_{2}}{{\alpha }_{2}}\Delta T$
We have already given the information that both rods are expanding by the same amount at the same temperature. So, we can equate change in length of each rod.
${{l}_{1}}{{\alpha }_{1}}\Delta T={{l}_{2}}{{\alpha }_{2}}\Delta T$
Since the temperature is same, we can rewrite this equation as,
${{l}_{1}}{{\alpha }_{1}}={{l}_{2}}{{\alpha }_{2}}$
$\dfrac{{{l}_{1}}}{{{l}_{2}}}=\dfrac{{{\alpha }_{2}}}{{{\alpha }_{1}}}$………………………..(1)
We can alter this equation by adding 1 to both sides.
$1+\dfrac{{{\alpha }_{1}}}{{{\alpha }_{2}}}=1+\dfrac{{{l}_{2}}}{{{l}_{1}}}$
$\dfrac{{{\alpha }_{2}}+{{\alpha }_{1}}}{{{\alpha }_{2}}}=\dfrac{{{l}_{1}}+{{l}_{2}}}{{{l}_{1}}}$
Or
$\dfrac{{{\alpha }_{2}}}{{{\alpha }_{1}}+{{\alpha }_{2}}}=\dfrac{{{l}_{1}}}{{{l}_{1}}+{{l}_{2}}}$
Therefore, the correct answer is A.

Additional information:
The coefficient of linear expansion can be written as,
${{\alpha }_{l}}=\dfrac{dL}{dT}$, where ${{\alpha }_{l}}$ is the coefficient of linear expansion, dL is the change in length and dT is the change in temperature.
The SI unit of coefficient of linear expansion is ${{K}^{-1}}$.
The cohesive forces between the atoms are the reason behind the expansion. It will alter according to the change in cohesive forces. For the less expansion, the cohesive forces have to be higher. Lead can expand suddenly even if the temperature changes slightly. Normally these types of materials won’t be used for building construction. For that purpose, we need highly stable metals. That’s why we are preferring metal alloys to get the higher linear coefficient of expansion.

Note: If we are doing this problem by taking the combined length initially, it will be difficult to find the right answer. So, it is better to consider each metal rod first, then we can combine by altering the equation. It is preferred to save time. We can simply guess the answer also. Since the length is in an inverse relationship with the temperature.