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A bar magnet is hung by a thin cotton th...

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 is

A

`(sqrt(3)W)/(2)`

B

`(2W)/(sqrt(3))`

C

`(W)/(sqrt(3))`

D

`sqrt(3)W`

Text Solution

Verified by Experts

The correct Answer is:
D
We know that
`tau = M xx B rArr tau = MB sin theta`
`tau = MB sin 60^(@)" "...(1)`
`W = MB(1 - cos 60^(@))" "...(2)`
From Eqs. (1) and (2)
`(tau)/(W) = (sqrt(3)//2)/(1//2) rArr tau = W sqrt(3)`
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