Choose the correct stability order of the given free radicals.

In compounds (I), (III) and (IV) the -M effect and hyperconjugation effect are inoperative at meta positions. So we should consider the inductive effect of the groups. -I effect will decrease the stability of the free radical by withdrawing the electrons toward it itself. The order of -I effect is (\mathrm{NO}_{2} > \mathrm{CN} > \mathrm{OH}). Therefore (III) is most destabilised. In compound (II) the OH act as electron donating group through +M effect which is operative at para position. This increase the electron density at para position and stabilise the free radical. Hence compound (1) is most stabilised. The order of stability of free radical is $\text{ II } > \text{ I } > \text{ IV } >$$\text{ III }$