A uniform wooden stick of mass 1.6 kg and length l rests in an inclined mannar on a smooth, vertical wall of height `h(ltl)` such that a small portion of the stick extends beyond the wall. The reaction force of th wall on the stick is perpendicular to the stick. The stick makes an angle of `30^@` with the wall and the bottom of the stick is on a rough floor. The reaction of the wall on the stick is equal in magnitude to the reaction of the floor on the stick. The ratio h/l and the friectional force f at the bottom of the stick are `(g = 10 ms^2)`

Answer

D

Explanation

Forces acting on the stick are shown in figure.
Normal reaction at point B acts perpendicular to the stick and we can resolve it along the horizontal and vertical direction as shown in figure.

Vertical equlilibrium:
`N+Nsin30^(@)=mg`
`(3)/(2)N=mg`
`N=(2)/(3)mg`
Horizontal equilibirum:
`f_(r)=Ncos30^(@)`
Substituting value of N we get
`f_(r)=(mg)/(sqrt3)=(16)/(sqrt3)=(16sqrt3)/(3)N`
Rotational equilibrium about point A:
`Nxx(h)/(cos30^(@))=mgxx(l)/(2)sin30^(@)`
`(2)/(3)mgxx(2h)/(sqrt3)=mgxx(1)/(4)`
`(h)/(l)=(3sqrt3)/(16)`
We can see that option (d) is correct.