The length of a wire becomes $$l_1$$ and $$l_2$$ when 100 N and 120 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 _______

Let the natural length of the wire be $$L$$ and its spring constant (stiffness) be $$k$$.

When tension of 100 N is applied:

$$l_1 = L + \frac{100}{k}$$ ... (i)

When tension of 120 N is applied:

$$l_2 = L + \frac{120}{k}$$ ... (ii)

We are given that $$10l_2 = 11l_1$$

Substituting:

$$10\left(L + \frac{120}{k}\right) = 11\left(L + \frac{100}{k}\right)$$

$$10L + \frac{1200}{k} = 11L + \frac{1100}{k}$$

$$\frac{1200}{k} - \frac{1100}{k} = 11L - 10L$$

$$\frac{100}{k} = L$$

Substituting back into equation (i):

$$l_1 = L + L = 2L$$

$$L = \frac{l_1}{2} = \frac{1}{2}l_1$$

Therefore, $$x = 2$$.