Correct Answer - C
`mg,(1)/(2)(sin60-sin30)=(1)/(2)kx^(2)`
`x=lcos30-lcos60`
`=mg,(1)/(2)(sqrt(3)/(2)-(1)/(2))=(1)/(2)kl^(2)(sqrt(3)/(2)-(1)/(2))^(2)`
`= (2mg)/(l(sqrt(3-1)))=k`
`tau_(A)=0implies mg ,(l)/(2)costheta=N_(1)lsintheta`
`N_(1)=(mgcottheta)/(2)`
`N_(1) = kx implies x=(mgcottheta)/(2k)`
`x=(mg cot 60^(@))/(2,(2mg)/(1.(sqrt(3-1))))=(sqrt(3-1))/(4sqrt(3))l`