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^(@) is W. Now the torrue required to keep the magnet in this new position isA. `(W)/(sqrt(3))`B. `sqrt(3)W`C. `(sqrt(3)W)/(2)`D. `(2W)/(sqrt(3))`

A bar magnet is hung by a thin cotton thread in a uniform horizontal m

1.	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^(@) is W. Now the torrue required to keep the magnet in this new position isA. `(W)/(sqrt(3))`B. `sqrt(3)W`C. `(sqrt(3)W)/(2)`D. `(2W)/(sqrt(3))`
Answer» Correct Answer - B `W=MB(cos^(@)-cos60^(@))` `W=MB(1-(1)/(2))=(MB)/(2)` required torque for this position `tau=Mbsintheta` `=MBsin60^@` `=(sqrt(3))/(2)MB=sqrt(3)W`

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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^(@) is W. Now the torrue required to keep the magnet in this new position isA. `(W)/(sqrt(3))`B. `sqrt(3)W`C. `(sqrt(3)W)/(2)`D. `(2W)/(sqrt(3))`

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