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Q.

The work done in turning a magnet of magnetic moment M by an angle of 90ofrom the meridian is n times the coresponding work done to turn it through an angle of 60o. Here n is

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a

½

b

2

c

¼

d

1

answer is B.

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Detailed Solution

W1=MB(cos0o-cos90o)=MB(1-0)=MBW2=MB(cos0o-cos60o)=MB(1-12)=MB2∴W1=2W2⇒n=2
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