A magnetic needle lying parallel to a magnetic field requires `W units` of work to turn it through `60^(@)`. The torque needed to maintain the needle in this position will be

Explanation

In case of a dipole in a magnetic field,
`W=MB(costheta_(1)-costheta_(2))andC=MBsintheta`
Here, `theta_(1)=0^(@)andtheta_(2)=60^(@)`
So, `W=MB(1-costheta)=2MB"sin"^(2)(theta)/2`
and, `C=MBsintheta=2MB"sin"(theta)/2"cos"theta/2`
So, `C/W=cot(theta/2),i.e" "C=Wcot30^(@)=sqrt3W`