A magnet of magnetic moment M is lying in a magnetic field of inclination B. W₁ is the work done in turning it from 0^o to 60^o and W₂ is the work done in turning it from 30^o to 90^o. Then

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W₂= W₁/2

W₂= W₁

W₂= 2W₁

W₂= $\sqrt{3}$ W₁

answer is D.

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Detailed Solution

$\begin{array}{l} W = MB (\cos θ_{1} - \cos θ_{2}) \\ ∴ W_{1} = MB (\cos 0^{\circ} - \cos 60^{\circ}) = mB / 2 \\ W_{2} = MB (\cos 30^{\circ} - \cos 90^{\circ}) = \sqrt{3} mB / 2 \\ ∴ W_{2} = \sqrt{3} W_{1} \end{array}$

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