The length of a wire becomes $l_1$ and $l_2$ when $100 \mathrm{N}$ and $120 \mathrm{N}$ tension are applied respectively. If $10l_2=11l_1$, then the natural length of wire will be $\frac{1}{x}{l}_1$. Here the value of $x$ is

Young's modulus is given by $Y=\frac{{\displaystyle \frac{F}{A}}}{{\displaystyle \frac{∆l}{l}}}$

$$\begin{gathered} \Rightarrow \frac{F}{A}=Y\left(\frac{\Delta l}{l}\right) \\ \Rightarrow \mathrm{\Delta l}=\frac{Fl}{AY} \end{gathered}$$

$$\begin{gathered} \Rightarrow l+\mathrm{\Delta l}=l_1 \\ \Rightarrow \left(\frac{F_{1}L}{AY}+L\right)=L\left(\frac{F_1}{AY}+1\right)=l_1 ...\left(i\right) \end{gathered}$$$$

Similarly,

$l_2=\left(\frac{F_2}{AY}+1\right)L$

$\Rightarrow 10l_2=11l_1$

It is given that $10l_1=11l_2$

$\Rightarrow 10\times L\left[\frac{F_2}{AY}+1\right]=11L\left[\frac{F_1}{AY}+1\right]$

$\Rightarrow \frac{10F_2}{AY}+10=\frac{11F_1}{AY}+11$

Substituting $F_1=100 \mathrm{N}, F_{\mathit{2}}=120 \mathrm{N}$

$$\begin{gathered} \Rightarrow \frac{1200}{AY}-\frac{1100}{AY}=1 \\ \Rightarrow \frac{100}{AY}=1 \end{gathered}$$

$\Rightarrow \frac{1}{AY}=\frac{1}{100}$

Using equation (i),

$L=\frac{l_1}{\left(\frac{F_1}{AY}+1\right)}=\frac{l_1}{\frac{100}{100}+1}=\frac{l_1}{2}$