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. ` 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 ` )
In first case, ` theta _ 1 = 0 and theta _ 2 = 90 ^@ `
` W _ 1 = MB ( cos 0 ^@ - cos 90 ^@) = MB `
In second case ` theta _ 1 = 0 ^@, theta _ 2 =60 ^@`
` rArr W _ 2 = MB ( cos 0 ^@ - cos 60 ^@) `
` W _ 2 = MB( 1 - ( 1 ) / ( 2 )) = ( MB) /( 2 ) `
Given, ` W _ 1 i = n W _ 2 `
` therefore MB = n ( MB) / ( 2 ) `
` rArr n = 2 `