Question

Two rods having length l${}_1$, and l${}_2$ made of materials with the linear expansion coefficient ${\alpha }_1$ and ${\alpha }_2$, were soldered together. The equivalent coefficients of linear expansion for the obtained rod :

• A. $\frac{l_1{\alpha }_2+l_2{\alpha }_1}{l_1+l_2}$
• B. $\frac{l_1{\alpha }_1+l_2{\alpha }_2}{{\alpha }_1+{\alpha }_2}$
• C. $\frac{l_1{\alpha }_1+l_2{\alpha }_2}{l_1+l_2}$
• D. $\frac{l_2{\alpha }_1+l_1{\alpha }_2}{{\alpha }_1+{\alpha }_2}$

Answer 1

Answer: (C)

$\frac{l_1{\alpha }_1+l_2{\alpha }_2}{l_1+l_2}$

The coefficient of linear expansion of first rod of length $l_1$ is:

${\alpha }_1=\frac{\Delta l_1}{l_1\times \Delta t}$

$\Delta l_1={\alpha }_1{l}_1\Delta t$ ............eq1

The coefficient of linear expansion of second rod of length $l_2$ is:

${\alpha }_2=\frac{\Delta l_2}{l_2\times \Delta t}$

$\Delta l_2={\alpha }_2{l}_2\Delta t$ ............eq2

Now, if a single rod of linear expansion $\alpha$ is taken instead of two rods,

$\alpha =\frac{\Delta l_1+\Delta l_2}{(l_1+l_2)\times \Delta t}$ ........eq3

Putting the value of $\Delta l_1$ and $\Delta l_2$ in eq3, we get

$\alpha =\frac{{\alpha }_1{l}_1\Delta t+{\alpha }_2{l}_2\Delta t}{(l_1+l_2)\Delta t}$

$\alpha =\frac{l_1{\alpha }_1+l_2{\alpha }_2}{l_1+l_2}$