A bar magnet of magnetic moment M 1 is suspended freely in a magnetic field through a wire. Initially there is no strain in wire. When top of the wire is twisted through 18 0 ∘ then the magnet rotat

Question

A bar magnet of magnetic moment M 1 ​ is suspended freely in a magnetic field through a wire. Initially there is no strain in wire. When top of the wire is twisted through 18 0 ∘ then the magnet rotates through 4 5 ∘ . If, under similar conditions, another magnet of magnetic moment M 2 ​ rotates through 3 0 ∘ then the ratio M 1 ​ : M 2 ​ is -

TestHub · Accepted Answer

Twist in wire balances magnetic torque: $C (ϕ - θ) = MB sin θ$

Magnet 1: $C (18 0^{\circ} - 4 5^{\circ}) = M_{1} B sin 4 5^{\circ}$

Magnet 2: $C (18 0^{\circ} - 3 0^{\circ}) = M_{2} B sin 3 0^{\circ}$

$\frac{M _{1}}{M _{2}} = \frac{135}{150} \cdot \frac{s i n 3 0 ^{\circ}}{s i n 4 5 ^{\circ}} = \frac{9}{10} \cdot \frac{1}{2} = \frac{9}{10 2}$

Answer: A

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JEE Main 2027 (Class 12)