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

Options

A. `(sqrt3W)/(2)`
B. `(2W)/(sqrt3)`
C. `(W)/(sqrt3)`
D. `sqrt3 W`

Answer

D

Explanation

`tau = PE sin 60^(@)`
`W= PE (1- cos 60^(@))`
From (i) and (ii)
`(tau)/(W)= (sqrt3//2)/(1//2) rArr tau = W sqrt3`