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A bar magnet is suspended by a string in a uniform magnetic field making an angle of $60^{o}$ with the field. Torque required to hold the magnet in that position is $25 \times 10^{- 3} N - m$ . Then the work done by an external agent to rotate the magnet clockwise through an angle of $90^{o}$ is (Take $\sqrt{3} = 1.7$ )

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1.50 J

zero

0.25 J

0.01 J

answer is D.

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Detailed Solution

$τ = M B \sin 60^{o} \Rightarrow M B = \frac{2 τ}{\sqrt{3}}$

Now $U_{i} = - M B \cos 60^{o} = - \frac{τ}{\sqrt{3}} a n d U_{f} = - M B (90^{o} - 60^{o}) = - τ$

$∴ W = U_{f} - U_{i} = = - τ - (- \frac{τ}{\sqrt{3}}) = - (\frac{\sqrt{3} - 1}{\sqrt{3}}) τ = 0.01 J$

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