The length of a metal wire is $l_1$ when the tension in it is $F_1$ and $l_2$ when the tension is $F_2$.…
The length of a metal wire is $l_1$ when the tension in it is $F_1$ and $l_2$ when the tension is $F_2$. Then, original length of the wire is
$\frac{l_1 F_1+l_2 F_2}{F_1+F_2}$
$\frac{l_2-l_1}{F_2-F_1}$
$\frac{l_1 F_2-l_2 F_1}{F_2-F_1}$
$\frac{l_1 F_1-l_2 F_2}{F_2-F_1}$
Solution
We have, $F_1 \propto$
$\left(l_1-l\right)$, where $\ell=$ original length Similarly, $F_2 \propto \quad\left(l_2-l\right)$, The ratio, $\frac{F_1}{F_2}=\frac{l_1-l}{l_2-l}$ Solving, we get
$\Rightarrow \quad l=\frac{F_2 l_1-F_1 l_2}{F_2-F_1}$