Magnetism Question 188
Question: 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
[CBSE PMT 1995; MP PET 2003]
Options:
A) 1/2
B) 2
C) 1/4
D) 1
Show Answer
Answer:
Correct Answer: B
Solution:
$ W _{1}=MB(\cos 0^{o}-\cos 90^{o})=MB(1-0)=MB $
$ W _{2}=MB(\cos 0^{o}-\cos 60^{o})=MB( 1-\frac{1}{2} )=\frac{MB}{2} $
$ \
Therefore \ W _{1}=2W _{2}\Rightarrow n=2 $
BETA
