A metallic bar is heated from $0^{\circ }\mathrm{C}\text{ to }{100}^{\circ }\mathrm{C}\text{. }$ Let the coefficient of linear expansion be ${10}^{-5}{\mathrm{K}}^{-1}\text{. }$The percentage increase in length must be :

As we know that, solid when expanded in one dimension (linear expansion) follows the following equation:

$l=l_o(1+\alpha \mathrm{\Delta }T)$

where ΔT is the change in temperature to cause the expansion. Expanding the above equation, we can get:

$\begin{array}{l}l-l_o=l_o\alpha \mathrm{\Delta }T\Rightarrow \mathrm{\Delta }l=l_o\alpha \\ \mathrm{\Delta }T\end{array}$

Now, from the question we have

$\begin{array}{l}\alpha ={10}^{-5}{\mathrm{K}}^{-1},\mathrm{\Delta }T\\ =\left({100}^{\circ }\mathrm{C}-0^{\circ }\mathrm{C}\right){100}^{\circ }\mathrm{C}\end{array}$

Putting these values in the above equation, we get

$\begin{array}{l}\Rightarrow \mathrm{\Delta }l=l_o\times {10}^{-5}\times 100\\ \Rightarrow \frac{\mathrm{\Delta }l}{l_o}={10}^{-3}\end{array}$

In percentage, $\frac{\mathrm{\Delta }l}{l_o}\times 100=0.1\mathrm{\%}$