A magnetic needle suspended parallel to a magnetic field requires $\sqrt 3\,J $ of work to turn it through $60^o$. The torque needed to maintain the needle in this position will be
  • A$\sqrt 3\,J $
  • B$\frac{3}{2}\,J$
  • C$2\sqrt {3\,} \,J$
  • D$3\,J$
Medium
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