A bar magnet is hung by a thin cotton thread in a uniform horizontal magnetic field and is in an equilibrium state. The energy required to rotate it by 60° is W. Now the torque required to keep the magnet in this new position isa) \(\frac{W}{\sqrt{3}}\)b) \(\frac{\sqrt{3}}{2}W\)c) \(\sqrt{3}W\)d) \(\frac{2W}{\sqrt{3}}\)

Option c) \(\sqrt{3}W\).

Was this answer helpful?

  
   

0 (0)

Upvote (0)

Choose An Option That Best Describes Your Problem

Thank you. Your Feedback will Help us Serve you better.

Leave a Comment

Your Mobile number and Email id will not be published. Required fields are marked *

*

*

BOOK

Free Class

Ask
Question