A magnet is suspended in the magnetic meridian with an untwisted wire. The upper end of wire is rotated through ${180}^{\mathrm{o}}$ to deflect the magnet by ${30}^{\mathrm{o}}$ from magnetic meridian. When this magnet is replaced by another magnet, the upper end of wire is rotated through ${270}^{\mathrm{o}}$ to deflect the magnet ${30}^{\mathrm{o}}$ from magnetic meridian. The ratio of magnetic moments of magnets is

detailed solution

Correct option is C

Let M1 and M2 be the magnetic moments of magnets and BH the horizontal component of earth's field. We have τ=MBH sinθ. if ϕ is the twist of wire, then τ=Cϕ, C being restoring couple per unit twist of wire⇒ Cϕ= MH sinθHere, ϕ1=180o-30o=150o=150 x π180 radϕ2=270o-30o=240o=240 x π180 radSo, Cϕ1 = M1BH sin θ (For deflection θ=30o of I magnet) Cϕ2 = M2BH sin θ (For deflection θ=30o of II magnet)⇒ϕ1ϕ2=M1M2⇒M1M2=ϕ1ϕ2=150 x π180240 x π180=1524=58⇒M1 : M2=5 : 8