The work done in turning a magnet of magnetic moment 'M' by an angle of `90^(@)` from the meridian is 'n' times the corresponding work done to turn it through an angle of `60^(@)`, where 'n' is given by

Options

A. 2
B. 1
C. `0.5`
D. `0.25`

Answer

A

Explanation

The work done in turning a magnet of magnetic moment M, from the magnetic meridian `(theta=0^(@))` to an angle `theta=90^(@)` given by
`W_(1)=MB[cos 0^(@) - cos 90^(@)]=MB[1-0]=MB`
and the work done in turning it through `60^(@)` is
`W_(2)=MB [cos 0^(@)-cos 60^(@)]`
`W_(2) =MB [1-1/2]=(MB)/2`
`:. W_(1)=2 W_(2)` But it is given that `W_(1)=n W_(2)`
Thus, `n=2`