A) \[W\]
B) \[2\,W\]
C) \[\sqrt{3}W\]
D) \[\frac{\sqrt{3}W}{2}\]
Correct Answer: C
Solution :
Work done in rotating a magnetic needle \[W=MB[\cos {{\theta }_{i}}-\cos {{\theta }_{f}}]\] Here, \[{{\theta }_{i}}=0\]and\[{{\theta }_{f}}=60{}^\circ \] \[W=MB[\cos 0{}^\circ -\cos 60{}^\circ ]\] \[=MB\left[ 1-\frac{1}{2} \right]\] Or \[W=\frac{MB}{2}\] ...(i) The torque required to maintain the magnet at same angle \[T=MB\text{ }sin\text{ }\theta =MB\text{ }sin\text{ }60{}^\circ \] or \[T=MB\left( \frac{\sqrt{3}}{2} \right)\] ...(ii) Dividing Eq. (ii) by Eq. (i), we get \[\frac{T}{W}=\sqrt{3}\] or \[T=W\sqrt{3}\]
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