1. Set up the situation

A mass $m$ is attached to a spring of natural length $l$ and spring constant $k$. One end is fixed at the centre of a horizontal rotating disc. When the disc rotates with angular speed $\omega$, the mass stays at rest relative to the disc at some stretched length.

Let the extension of the spring be $x$.

Then the new length of the spring is $r=l+x.$

The spring force provides the required centripetal force.

2. Write the force balance

Spring force: $F_s = kx$

Required centripetal force for circular motion of radius $r$: $F_c = m\omega^2 r = m\omega^2(l+x)$

So, $kx = m\omega^2(l+x).$

Rearranging, $kx - m\omega^2x = m\omega^2 l$ $x(k-m\omega^2)=m\omega^2 l$

Hence, $x = \frac{m\omega^2 l}{k-m\omega^2}.$

3. Use the given approximation

Given: $k \gg m\omega^2$

Therefore, $k-m\omega^2 \approx k.$

So, $x \approx \frac{m\omega^2 l}{k}.$

The relative change in length is $\frac{x}{l} \approx \frac{m\omega^2}{k}.$

4. Match with options

The relative change in length is best given by $\boxed{\frac{m\omega^2}{k}}$

So the correct option is B.

5. Check against stored answer

Stored correct answer: B

This matches our derived answer.