Verified Solution • 2025–26 Session moderate

QAIPM-2012-Ph-1651: NEET 2012 Physics - Class 12 AIPM-2012-Ph-1651 Solution: A magnetic needle suspended parallel to a magnetic field…

A magnetic needle suspended parallel to a magnetic field requires 3 J\sqrt{3}\ \text{J} of work to turn it through 60∘60^\circ. The torque needed to maintain the needle in this position will be 23 J2\sqrt{3}\ \text{J} 3 J 3 J\sqrt{3}\ \text{J} 23 J2\sqrt{3}\ \text{J}

Multiple Choice Options

  • A 23J2\sqrt 3 J 2 3 _ J
  • B 33 3 JJ J
  • C 3 J\sqrt 3 \,J 3 _ J
  • D 32 J{3 \over 2}\,J 2 3 _ J

Step-by-Step Verified Solution

According to work energy theorem W = U final � U initial = MB (cos 0 � cos 60�) = M B 2 = 3 J ....(1) The torque on the needle is _ _ = M _ _ B _ = M B sin _ _ = M B sin _ 60 _ = M B 3 2 ....(2) From equation (1) and (2), _ = 2 3 _ 3 2 = 3 J

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