A bar magnet of magnetic moment M1  is suspended by a wire in a magnetic field. The upper end of  the wire is rotated through 1800, then the magnet rotated through 450.under similar conditions another magnet of magnetic moment M2 is rotated through 300. Then find the ratio of  M1 and M2

A magnet of  moment  'M' is suspended in the magnetic  meridian with an un twisted wire.The upper  end of  the wire is rotated through an angle  α  to  deflect the magnet by an angle θ from magnetic meridian. Then deflecting couple acting on the magnet = mBHsinθRestoring couple developed in suspension wire =  C(α−θ)C  is  the couple per unit twist of suspension wire.In equilibrium position,   mBHsinθ=C(α−θ) Given  θ1=450   θ2=30°α=180°  for first magnet  C(180-45)=M1sin45°for second magnet  C(180-30)=M2sin30°                ⇒135150=M1M2sin45°sin30°⇒135150=M1M212X2⇒135150=M1M2(2)⇒M1M2=9102