A) 1
B) ¼
C) 4
D) 2
Correct Answer: D
Solution :
If dipole be rotated from an initial orientation \[\theta ={{\theta }_{1}}\]to final orientation \[\theta ={{\theta }_{2}},\] the total work required will be \[W=\int\limits_{{{\theta }_{1}}}^{{{\theta }_{2}}}{MB\sin \theta d\theta }\] \[=MB[-cos\theta ]_{{{\theta }_{1}}}^{{{\theta }_{2}}}\] \[W=MB(cos{{\theta }_{1}}-cos{{\theta }_{2}})\] First case, \[{{\theta }_{1}}=0\]and \[{{\theta }_{2}}={{90}^{o}}\] \[\therefore \] \[{{W}_{1}}=MB(cos\theta -cos{{90}^{o}})=MB\] Second case, \[{{\theta }_{1}}=0\]and \[{{\theta }_{2}}={{60}^{o}}\] \[\therefore \] \[{{W}_{2}}=MB(cos\theta -cos{{60}^{o}})\] \[=MB\left( 1-\frac{1}{2} \right)=\frac{MB}{2}\] Given, \[{{W}_{1}}=n{{W}_{2}}\] \[\therefore \] \[MB=n\frac{MB}{2}\] \[\Rightarrow \] \[n=2\]
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