When the tension on a wire is 4N its length is l₁. When the tension on the wire is 5N its length is l₂. Find its natural length.

T₁ = F₁ = 4N $\to$l₁ ; T₂ = F₂ = 5N $\to$l₂ 
Let l be the natural, original or initial length e $\propto$F
$\therefore \frac{{\mathrm{e}}_1}{{\mathrm{e}}_2}=\frac{{\mathrm{F}}_1}{{\mathrm{F}}_2}\Rightarrow \frac{{\mathrm{l}}_1-\mathrm{l}}{{\mathrm{l}}_2-\mathrm{l}}=\frac{4}{5}$

$5{\mathrm{l}}_1-5\mathrm{l}=4{\mathrm{l}}_2-4\mathrm{l}\Rightarrow \mathrm{l}=5{\mathrm{l}}_1-4{\mathrm{l}}_2$