A small mass attached to a string rotates on a frictionless table top as shown. If the tension in the string is decreased by relasing the string causing the radius of the circular motion to increase by a factor of $2$, the kinetic energy of the mass will :- 

As the string is pulled downwards, tension in the string, always pass through point $O$. 

So, torque of tension about $O$ will be zero and angular momentum of object about $O$ will remain constant. 

also, $K=\frac{L^2}{2I}$ 

$\Rightarrow K\propto \frac{1}{I}\propto \frac{1}{m{r}^2}$           $\left\{\therefore L is cons\tan t\right\}$

so, if $r$ becomes $2r$ then $K$ becomes $\frac{K}{4}$.