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. `1//2`
B. 2
C. `1//4`
D. 1

Answer

A

Explanation

`W_(1)= - MB (cos 90^(@)- cos 0^(@)) = MB` ...(i)
`W_(2)= - MB (cos 60^(@)- cos 0^(@))= (MB)/(2)` ...(ii)
By equation (i) and (ii)
`n= (1)/(2) ( because W_(1)= n W_(2))`