Explanation

(c) Let $M_{1}$ and $M_{2}$ be the magnetic moments of magnets and $H$ the horizontal component of earth’s field. We have $\tau = MH\sin\theta$. If $\phi$ is the twist of wire, then $\tau = C\phi$, C being restoring couple per unit twist of wire
$\Rightarrow C\phi = MH\sin\theta$
Here $\phi_{1} = (180{^\circ} - 30{^\circ}) = 150{^\circ} \times \frac{\pi}{180}rad$
$\phi_{2} = (270{^\circ} - 30{^\circ}) = 240{^\circ} = 240 \times \frac{\pi}{180}rad$
So, $C\phi_{1} = M_{1}H\sin\theta$ [For deflection $\theta = 30{^\circ}$ of I magnet]
$C\phi_{2} = M_{2}H\sin\theta$ [For deflection $\theta = 30{^\circ}$ of II magnet]
Dividing $\frac{\phi_{1}}{\phi_{2}} = \frac{M_{1}}{M_{2}}$
$\Rightarrow \frac{M_{1}}{M_{2}} = \frac{\phi_{1}}{\phi_{2}} = \frac{150 \times \left( \frac{\pi}{180} \right)}{240 \times \left( \frac{\pi}{180} \right)} = \frac{15}{24} = \frac{5}{8}$
$\Rightarrow M_{1}:M_{2} = 5\ :8$