Two different wires having lengths ${\mathrm{L}}_1$ and ${\mathrm{L}}_2$, and respective temperature coefficient of linear expansion ${\mathrm{ɑ}}_1$ and ${\mathrm{ɑ}}_2$, are joined end-to-end. Then the effective temperature coefficient of linear expansion is:

The correct option is A

$[{\mathrm{ɑ}}_1{\mathrm{L}}_1+{\mathrm{ɑ}}_2{\mathrm{L}}_2]/[{\mathrm{L}}_1+{\mathrm{L}}_2]$

Explanation for correct option:

Option (a) $[{\mathrm{ɑ}}_1{\mathrm{L}}_1+{\mathrm{ɑ}}_2{\mathrm{L}}_2]/[{\mathrm{L}}_1+{\mathrm{L}}_2]$

Step 1: Given data

Temperature coefficient of linear expansion for wire ${\mathrm{L}}_1$= ${\mathrm{ɑ}}_1$

Temperature coefficient of linear expansion for wire ${\mathrm{L}}_2$= ${\mathrm{ɑ}}_2$

Step 2: Formula used

The coefficient of Linear Expansion is the rate of change of unit length per unit degree change in temperature

The coefficient of linear expansion can be mathematically written as:

${\mathrm{\alpha }}_{\mathrm{L}}=\frac{\mathrm{dL}}{\mathrm{dT}}$

Where, ${\mathrm{\alpha }}_{\mathrm{L}}$is the coefficient of linear expansion, $\mathrm{dL}$is the unit change in length, and $\mathrm{dT}$ is the unit change in temperature.

Step 3: Find the effective temperature coefficient of linear expansion

The coefficient of linear expansion of the first wire of length ${\mathrm{L}}_1$ is

${\mathrm{\Delta L}}_1={\mathrm{\alpha }}_1{\mathrm{L}}_1\mathrm{\Delta t}$

and, the coefficient of linear expansion of the second wire of length ${\mathrm{L}}_2$ is

${\mathrm{\Delta L}}_2={\mathrm{\alpha }}_2{\mathrm{L}}_2\mathrm{\Delta t}$

Now, if a single wire of linear expansion$\mathrm{\alpha }$ is taken instead of two wires,

$\mathrm{\alpha }=\frac{{\mathrm{\Delta L}}_1+{\mathrm{\Delta L}}_2}{{\mathrm{L}}_1+{\mathrm{L}}_2\times \mathrm{\Delta t}}$

Putting the value of$∆{\mathrm{L}}_1$ and $∆{\mathrm{L}}_2$ , we get:

$\mathrm{\alpha }=\frac{{\mathrm{\alpha }}_1{\mathrm{L}}_1+{\mathrm{\alpha }}_2{\mathrm{L}}_2}{{\mathrm{L}}_1+{\mathrm{L}}_2}$

Thus, the effective temperature coefficient of linear expansion is $\left(\frac{{\mathrm{\alpha }}_1{\mathrm{L}}_1+{\mathrm{\alpha }}_2{\mathrm{L}}_2}{{\mathrm{L}}_1+{\mathrm{L}}_2}\right)$

So, the correct option is (a).