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Physics · PYQ NEET UG 2016(2)
हिन्दीA bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by \( 60^\circ \) is W. Now the torque required to keep the magnet in this new position is
Correct answer · Option 3
\( \sqrt{3} W \)
Explanation
Work done \( W = MB(\cos 0^\circ - \cos 60^\circ) = MB(1 - 1/2) = MB/2 \). Torque at \( 60^\circ \) is \( \tau = MB \sin 60^\circ = (2W) \frac{\sqrt{3}}{2} = \sqrt{3}W \).
