Since the pulling force (tension) acts along the string towards the center of rotation, the torque about the center of the circular path is zero ($\tau = \mathbf{r} \times \mathbf{F} = 0$). Therefore, the angular momentum ($L$) of the mass is conserved. $L = mvr = \text{constant}$ The kinetic energy $K$ of the particle can be written in terms of angular momentum as: $K = \frac{1}{2}mv^2 = \frac{1}{2}m\left(\frac{L}{mr}\right)^2 = \frac{L^2}{2mr^2}$ Since $L$ and $m$ are constants, we have $K \propto \frac{1}{r^2}$. If the radius $r$ decreases by a factor of $2$ (i.e., the new radius is $r' = r/2$), the new kinetic energy $K'$ becomes: $K' = \frac{L^2}{2m(r/2)^2} = 4 \left(\frac{L^2}{2mr^2}\right) = 4K$ Thus, the kinetic energy increases by a factor of $4$.