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

Let the original length be 'l₀'
When T₁ = 100N, Extension = l₁-l₀  
When T₂ = 120N, Extension = l₂-l₀ 
Then 100 = K(l₁-l₀)
and 120=K (l₂-l₀)
$\begin{array}{l}\Rightarrow \frac{5}{6}=\frac{{\mathrm{\ell }}_1-{\mathrm{\ell }}_0}{{\mathrm{\ell }}_2-{\mathrm{\ell }}_0}\\ \Rightarrow 5{\mathrm{\ell }}_2-5{\mathrm{\ell }}_0=6{\mathrm{\ell }}_1-6{\mathrm{\ell }}_0\\ \Rightarrow {\mathrm{\ell }}_0=6{\mathrm{\ell }}_1-5{\mathrm{\ell }}_2\\ \Rightarrow {\mathrm{\ell }}_0=6{\mathrm{\ell }}_0-5\left(\frac{11{\mathrm{\ell }}_1}{10}\right)\\ \Rightarrow {\mathrm{\ell }}_0=\frac{{\mathrm{\ell }}_1}{2}\therefore \mathrm{x}=2\end{array}$