A bar magnet of magnetic moment $M_1$  is suspended by a wire in a magnetic field. The upper end of  the wire is rotated through ${180}^0$, then the magnet rotated through ${45}^0$.under similar conditions another magnet of magnetic moment $M_2$ is rotated through ${30}^0$. Then find the ratio of  $M_1$ and $M_2$ 

A magnet of  moment  $\text{'}M\text{'}$ is suspended in the magnetic  meridian with an un twisted wire.

The upper  end of  the wire is rotated through an angle  $\alpha$  to  deflect the magnet by an angle $\theta$ from magnetic meridian. Then deflecting couple acting on the magnet = $m{B}_H\sin \theta$

Restoring couple developed in suspension wire =  $C(\alpha -\theta )$

C  is  the couple per unit twist of suspension wire.

In equilibrium position,   $m{B}_H\sin \theta =C(\alpha -\theta )$
 Given  ${\theta }_1={45}^0$
   $\begin{array}{l}{\theta }_2=30°\\ \alpha =180°\end{array}$
  
for first magnet  $C(180-45)=M_1\sin 45°$
for second magnet  $C(180-30)=M_2\sin 30°$

                $\begin{array}{l}\Rightarrow \frac{135}{150}=\frac{M_1}{M_2}\frac{\sin 45°}{\sin 30°}\\ \Rightarrow \frac{135}{150}=\frac{M_1}{M_2}\frac{1}{\sqrt{2}}X2\\ \Rightarrow \frac{135}{150}=\frac{M_1}{M_2}\left(\sqrt{2}\right)\\ \Rightarrow \frac{M_1}{M_2}=\frac{9}{10\sqrt{2}}\end{array}$