Does the length be useful everytime !!!!

The extension of a spring is given by $\frac{F}{k} = \Delta x$ where $F = mg$ .

Therefore, $\frac{mg}{k} = \Delta x$ .

The weight acting on a particular point along the string can be given by $mgn$ where $n$ = how far up the string the point is, expressed as a fraction. Therefore, $0\leq n \leq 1$ .

0 is at the bottom of the string (no weight/force acts to extend it) and 1 is at the top of the string where the string's entire weight provides a force to extend it.

We want the extension of the whole string, so we can integrate the function between 0 and 1, because this interval for $n$ encompasses the string's whole length.

$\int_0^1 \frac{mgn}{k}\;\mathrm{d}n= \tfrac{mgn^{2}}{2k} \Big|_0^1 = \frac{mg}{2k} = \frac{1 \times 10}{2 \times 500} = \boxed{0.01m}$ .

I guess you could also just state that the 'average' weight (force) acting on the whole string is $\frac{mg}{2}$ , i.e. half the total weight, and substitute this directly into the force-extension equation to yield the same answer.