The workdone to trun a magnet by 60^(@) from its equilibrium position is W in a uniformthe torque required to hold it in that postion will be

The workdone to trun a magnet by 60^(@) from its equilibrium position

1.	The workdone to trun a magnet by 60^(@) from its equilibrium position is W in a uniformthe torque required to hold it in that postion will be
Answer» `(w)/(2)` `(w)/sqrt(3)` `sqrt(3)/(2)w` `wsqrt(3)` Solution :`W=MB =(mB)/(2)` `tau-mBsin 60^(@),tau=mBxxsqrt(3)/(2)=Wxxsqrt(3)`

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The workdone to trun a magnet by 60^(@) from its equilibrium position is W in a uniformthe torque required to hold it in that postion will be

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