Explain linear expansion. Write a unit.

Increase in length of substance $(\Delta l)$ is directly proportional to original length " $l$ " and increase in temperature "$\Delta T$".

$\therefore \Delta l \alpha l$ and $\Delta l \alpha \Delta \mathrm{T}$

$\therefore \Delta l \alpha l \Delta \mathrm{T}$ (combinely)

$\therefore \frac{\Delta l}{l} \alpha \Delta \mathrm{T}$

Hence, the fractional change in length. $\left(\frac{\Delta l}{l}\right)$ is directly proportional to $\Delta \mathrm{T}$.

$\therefore \frac{\Delta l}{l} \propto \Delta \mathrm{T}$

$\therefore \frac{\Delta l}{l}=\alpha_{l} \Delta \mathrm{T}$

$\therefore \Delta l=\alpha_{l} l \Delta \mathrm{T}$

Where $\alpha_{l}$ is coefficient of linear expansion and is characteristic of material.

Value of $\alpha_{l}$ depends on type of material and temperature.

If the temperature difference is not large then ' $\alpha_{l}$ ' doesn't depend on temperature.

Unit of $\alpha_{l}$ is $\left({ }^{\circ} \mathrm{C}\right)^{-1}$ or $\mathrm{K}^{-1}$.

In equation (1), $\Delta l=l_{2}-l_{1}$ and $\Delta \mathrm{T}=\mathrm{T}_{2}-\mathrm{T}_{1}$,

$l_{2}-l_{1}=\alpha_{l} l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)$

$\therefore l_{2}=l_{1}+\alpha_{l} l_{1}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)$

$\therefore l_{2}=l_{1}\left[1+\alpha_{l}\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right)\right]$

By taking $\alpha_{l}=\alpha$ and $l_{1}=l$

$l_{2}=l\left[1+\alpha\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)\right]$