A body of mass 200g is tied to a spring of spring constant 12 . 5N...

Let the extension in spring be $x$ and the natural length of spring be $l$.

Here the centrifugal force will stretch the spring.

So, centrifugal force=spring force

$kx = m\left(l + x\right){\omega}^{2} ⇒ 12 . 5x = \dfrac{1}{5}\left(l + x\right)25 ⇒ 1 . 5x = l ⇒ \dfrac{x}{l} = \dfrac{1}{1 . 5} = \dfrac{2}{3}$

$⇒ x: l = 2: 3$