Correct Answer - 6
For equilibrium of the cylinder in the horizontal direction
`N_(1)=muN_(2)`……….i
In the vertical direction
`N_(2)+muN_(1)=F+mg`…….ii
Solving these two equations with `F=40N, M=2kg` and `mu=1/3`, we get
`N_(1)=18N`
and `N_(2)=54N`
Now, `Fd=mu(N_(1)+N_(1))r`
`:. d=(M(N_(1)+N_(2))r)/F`
`=((1/3)(18+50)(0.1))/40=0.06m`