Given:

1. When tension $$T_1 = 100 \mathrm{~N}$$, extension $$= l_1 - l_0$$.
2. When tension $$T_2 = 120 \mathrm{~N}$$, extension $$= l_2 - l_0$$.

Now, let's write the equations using Hooke's Law:

$$100 = k(l_1 - l_0)$$

$$120 = k(l_2 - l_0)$$

Divide the first equation by the second equation:

$$\frac{5}{6} = \frac{l_1 - l_0}{l_2 - l_0}$$

Given the relationship between $$l_1$$ and $$l_2$$:

$$10l_2 = 11l_1$$

Now, let's solve for $$l_0$$:

$$5l_2 - 5l_0 = 6l_1 - 6l_0$$

$$l_0 = 6l_1 - 5l_2$$

Substitute the relationship between $$l_1$$ and $$l_2$$:

$$l_0 = 6l_1 - 5\left(\frac{11l_1}{10}\right)$$

$$l_0 = 6l_1 - \frac{11l_1}{2}$$

$$l_0 = \frac{l_1}{2}$$

Therefore, the natural length of the wire is $$\frac{1}{x}l_1 = \frac{2}{1}l_1 = 2l_1$$.

The value of $$x$$ is $$2$$.