A body of mass 200   g is tied to a spring of spring constant 12 . 5   N   m - 1 , while the…

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

$$\begin{gathered} kx=m\left(l+x\right){\omega }^2 \\ \Rightarrow 12.5x=\frac{1}{5}\left(l+x\right)25 \\ \Rightarrow 1.5x=l \\ \Rightarrow \frac{x}{l}=\frac{1}{1.5}=\frac{2}{3} \end{gathered}$$

$\Rightarrow x:l=2:3$