Question 115 of 210

A body of mass $200g$ is tied to a spring of spring constant $12.5N{m}^{-1}$, while the other end of spring is fixed at point $O$. If the body moves about $O$ in a circular path on a smooth horizontal surface with constant angular speed $5rad{s}^{-1}$, then the ratio of extension in the spring to its natural length will be :

A$1:2$B$1:1$C$2:3$D$2:5$

Show answer and solution▾

Answer: (C) $2:3$

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