Verified Solution • 2025–26 Session moderate

QNEET-2016-Ph-1168: NEET 2016 Physics - Class 12 NEET-2016-Ph-1168 Solution: A bar magnet is hung by a thin cotton thread in a uniform…

A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in equilibrium state. The energy required to rotate it by 60∘60^\circ is W. Now the torque required to keep the magnet in this new position is W3\frac{W}{\sqrt{3}} 3W\sqrt{3}W 3W2\frac{\sqrt{3}W}{2} 2W3\frac{2W}{\sqrt{3}}

Multiple Choice Options

  • A W3{W \over {\sqrt 3 }} 3 _ W _
  • B 3W\sqrt 3 W 3 _ W
  • C 3W2{{\sqrt 3 W} \over 2} 2 3 _ W _
  • D 2W3{{2W} \over {\sqrt 3 }} 3 _ 2 W _

Step-by-Step Verified Solution

W = MB (cos 0� � cos 60�) _ W = MB ( 1 _ 1 2 ) = M B 2 ....(1) Required torque for this position _ = MB sin _ = MB sin 60� = 3 2 M B = 3 W [From (1)]

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