A) \[\text{MH}\,\text{cos}\,\]
B) \[\text{2MH}\,{{\sin }^{2}}\frac{\theta }{2}\]
C) \[\text{MH}\,\sin \]
D) \[\text{MH}\,(1-\cos \text{)}\]
Correct Answer: D
Solution :
Work done in rotating the dipole \[W=MH(1-\cos \,)\]
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