Two magnets are suspended by a given wire one $45°,$ the wire has to be twisted through $540°$ whereas with the second magnet, the wire requires a twist of $360°$ for the same deflection. Then the ratio of magnetic moments of the two magnets is

If $C$ is torque per unit angular twist of the wire, then for a twist $\varphi$,

$\tau =C\varphi =MB\sin \theta .$

In the $1^{\text{st }}case, {\varphi }_1=540°-45°=495°, {\theta }_1=45°$

In the $2^{\text{nd }}case, {\varphi }_2=360°-45°=315°, {\theta }_2=45°$

$\therefore C\left(495°\right)=M_{1}B\sin 45° and C\left(315°\right)=M_{2}B\sin 45°$

Dividing $\left(i\right)$ by $\left(ii\right)$, we get $\frac{M_1}{M_2}=\frac{495}{315}=\frac{11}{7}$