Question

Two rods of length ${L_1}$ and ${L_2}$ are welded together to make a composite rod of length ${L_1} + {L_2}$. If the coefficient of linear expansion of the materials of the rods are ${\alpha _1}$ and ${\alpha _2}$ respectively, the effective coefficient of linear expansion of the composite rod is:
A. $\dfrac{{{L_1}{\alpha _1} - {L_2}{\alpha _2}}}{{{L_1} + {L_2}}}$
B. $\dfrac{{{L_1}{\alpha _1} + {L_2}{\alpha _2}}}{{{L_1} + {L_2}}}$
C. $\sqrt {{\alpha _1}{\alpha _2}}$
D. $\dfrac{{{\alpha _1} + {\alpha _2}}}{2}$

Answer 1

We can calculate this problem by using the coefficient of linear expansion formula. Remember that the change in length of the composite rod due to change in temperature is the sum of change in length of the individual rods due to the same amount of change in temperature.

Formula Used:
The change in length of a material because of linear expansion $\Delta l$ is given to be $\alpha L\Delta T$, where $L$ represents the original length of the material, $\alpha$ is the coefficient of linear expansion of material and $\Delta T$ is the change in temperature.

Complete step by step answer:
Linear expansion is the expansion of length of a material when subjected to an increase in temperature. Since we make the composite rod by simply welding the individual rods, we have its length $L = {L_1} + {L_2}$. Also note that the change in length in the composite rod due to a $\Delta T$ change in temperature, is equal to the sum of the change of length of the two rods.

Let $\alpha$ be the coefficient of linear expansion of the composite rod. Let $\Delta l$ represent the change in length of the composite rod, while $\Delta {l_1}$ and $\Delta {l_2}$ be the change in length of the individual rods due to a $\Delta T$ change in temperature. Then we have $\Delta l = \Delta {l_1} + \Delta {l_2}$. - - - - - - - - - - - - - - - (1)
Also, by using the linear expansion formula we have $\Delta l = \alpha L\Delta T$ - - - - - - - - - - - (2).
From (1) and (2) we have $\Delta {l_1} + \Delta {l_2} = \alpha L\Delta T$.

Now by using the formula of linear expansion again we have, $\Delta {l_1} = {\alpha _1}{L_1}\Delta T$ and $\Delta {l_2} = {\alpha _2}{L_2}\Delta T$.
$\Rightarrow {\alpha _1}{L_1}\Delta T + {\alpha _2}{L_2}\Delta T = \alpha L\Delta T$
$\Rightarrow {\alpha _1}{L_1} + {\alpha _2}{L_2} = \alpha L$
$\Rightarrow \alpha = \dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{L}$
$\therefore \alpha = \dfrac{{{\alpha _1}{L_1} + {\alpha _2}{L_2}}}{{{L_1} + {L_2}}}$

Hence the correct option is B. $\dfrac{{{L_1}{\alpha _1} + {L_2}{\alpha _2}}}{{{L_1} + {L_2}}}$.

Note: Thermal expansion is described as the tendency of matter to change its shape, area and volume in response to a change in temperature. Linear expansion denotes such change in one dimension (length) as opposed to change in volume (volumetric expansion).