Note: This question was given bonus by JEE council.

The magnetic field in a solenoid is given by, $\overrightarrow{B}={\mu }_{0}mI\hat{\mathrm{n}}$. Therefore, the flux through the loop will be,

$\varphi =\left(A\hat{k}\right)·\left({\mu }_{0}mI\hat{\mathrm{n}}\right)$

For case I:

$\varphi =\frac{BA}{\sqrt{2}}\cos \left(\omega t\right)$

The emf will be,

$\epsilon =\frac{BA\omega }{\sqrt{2}}\sin \left(\omega t\right)$

Therefore, the current in the loop will be,

$i=\frac{BA\omega }{\sqrt{2}R}\sin \left(\omega t\right)$

The magnetic moment of the loop will be,

$\overrightarrow{m}=iA\hat{\mathrm{k}}=\frac{B{A}^2\omega }{\sqrt{2}R}\sin \left(\omega t\right)\hat{\mathrm{k}}$

Therefore, the torque

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}=\frac{B^2{A}^2\omega }{\sqrt{2}R}\sin \left(\omega t\right)\left(\hat{\mathrm{k}}\times \hat{\mathrm{n}}\right)$

$\Rightarrow \overrightarrow{\tau }=-\frac{B^2{A}^2\omega }{2R}\left[\hat{\mathrm{i}}\right]{\sin }^2\left(\omega t\right)$

$\Rightarrow \overrightarrow{\tau }=-\frac{B^2{A}^2\omega }{2R}\left[{\sin }^2\left(\frac{\pi }{6}\right)\right]=\frac{-\alpha }{4}\hat{\mathrm{i}}$

Hence, $\left(\mathrm{I}\right)\to q$

For case II

$\varphi =0$

Therefore, we can directly write $\tau =0$

Hence $\left(\mathrm{II}\right)\to p$

Following the same process as in case I for case III

$\varphi =\frac{BA}{\sqrt{2}}\cos \left(\omega t\right)$

$i=\frac{BA\omega }{\sqrt{2}R}\sin \left(\omega t\right)$

$\overrightarrow{m}=\frac{B{A}^2\omega }{\sqrt{2}R}\sin \left(\omega t\right)\hat{\mathrm{k}}$

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}=\frac{B^2{A}^2\omega }{\sqrt{2}\times \sqrt{2}R}\sin \omega t\left(\hat{\mathrm{k}}\times \left(\sin \omega t\hat{\mathrm{i}}+\cos \omega t\hat{\mathrm{k}}\right)\right)$

$\overrightarrow{\tau }=\frac{B^2{A}^2\omega \sin \left(\omega t\right)}{2R}\sin \left(\omega t\right)\hat{\mathrm{j}}$

$\overrightarrow{\tau }=\frac{B^2{A}^2\omega }{2R}{\sin }^2\left(\omega t\right)\hat{\mathrm{j}}$

$\overrightarrow{\tau }=\frac{\alpha }{4}\hat{\mathrm{j}}$

Hence, $\mathrm{III}\to s$

Following the same process as in case I for case IV

$\varphi =\frac{BA}{\sqrt{2}}\sin \left(\omega t\right)$

$i=-\frac{BA\omega }{\sqrt{2}R}\cos \left(\omega t\right)$

$\overrightarrow{m}=-\frac{B{A}^2\omega }{\sqrt{2}R}\cos \omega t\left(\hat{\mathrm{k}}\right)$

$\overrightarrow{\tau }=\overrightarrow{m}\times \overrightarrow{B}=-\frac{B^2{A}^2\omega }{2R}\left(\hat{\mathrm{k}}\times \hat{\mathrm{j}}\right){\cos }^2\left(\omega t\right)$

$\overrightarrow{\tau }=+\frac{B^2{A}^2\omega }{2R}\left(\hat{\mathrm{i}}\right)·{\cos }^2\left(\frac{\pi }{6}\right)$

$\overrightarrow{\tau }=+\frac{3}{4}\alpha \hat{\mathrm{i}}$

Hence, $\left(\mathrm{IV}\right)\to r$