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Q.

A bar magnet of moment 2 A-m2 is free to rotate about a Vertical axis passing through its centre. The magnet is released form rest from east west direction. The K.E of the magnet as it takes north-south direction is (BH = 25 × 10–6 T)

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a

25 × 10–6 J

b

50 × 10–6 J

c

75 × 10–6 J

d

100 × 10–6 J

answer is B.

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Detailed Solution

∆KE=∆PE=MB(1-cosθ) =2 x 25 x 10-6(1-0)=50 x 10-6 J
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