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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°. The value of n is given by (a) 2 (b) 1 (c) 0.5 (d) 0.25



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°. The value of n is given by    


(a)    2    


(b)    1


(c)    0.5   


(d)    0.25


 


Grade:10

1 Answers

Vikas TU
14149 Points
11 months ago
Magnetic moment = M
Initial angle through which magnet is turned (q1) = 90°
Final angle through which magnet is turned (q2) = 60°.
Work done in turning the magnet, through 90° (W1) = MB (cos 0°-cos 90°)  = MB
Work done in turning the magnet, through 60°(W2) = MB (cos 0°-cos 60°)  = 0.5MB
= W1 = 2W2
= W1 = nw2 ( Given)
Thus n = 2
Therefore the corresponding work done to turn through an angle of 60° is 2
 
 

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