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Q.

The work done in turning a magnet of magnetic moment 'M' by an angle of 90° from the meridian is 'n' times the corresponding work done to turn it through an angle of 60°, where 'n' is given by

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a

1/2

b

2

c

1/4

d

1

answer is B.

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Detailed Solution

W1=MBcos⁡0∘−cos⁡90∘=MB(1−0)=MBW2=MBcos⁡0∘−cos⁡60∘=MB1−12=MB2∴W1=2W2⇒n=2
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