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∘ is

Options

A. n=1/2
B. n=2
C. n=1/4
D. n=1

Answer

B

Explanation

Work done, `W = MB(cos theta_(1) - cos theta_(2))` for rotating `60^(@)`
`W_(2) = MB(cos0^(@) - cos 60^(@))`
`=MB (1-1//2)`
`= (MB)/2`
For rotating `90^(@)`
`W_(1) = MB(cos 0^(@) - cos 90^(@))`
`W_(1) = MB(cos0^(@) - cos 90^(@))`
`W_(1) = MB`
Given, `W_(1) = nW_(2)`
`MB = n xx (MB)/2`
n=2