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

Answer

C

Explanation

Work done in turning the magnet through `60^(@)`
`(W_(1)) = MB(cos 0^(@) - cos 60^(@)) = MB(1- 1/2) =(MB)/2`
turning the magnet through `90^(@)`
`W_(2) = MB(cos 0^(@) - cos 90^(@)) = MB`
According to the equation
`W_(2) = nW_(1)`......(1)
`therefore` Here `W_(2)/W_(1) = (MB)/(MB//2) =2`
`W_(2) = 2W_(1)`.......(2)
Comparing (1) and (2)
n=2