Correct Answer - d
Here, `m = 1.6kg length =l(gt h)`
`N_(1)` is reaction of wall perpendicular to rod `N_(2)` is reaction of ground normal to floor `f` is force of friction along the floor towards the wall
In equilibrium `Sigma F_(x) =0`
`N_(1) cos30^(@)-f=0`
`Sigma F_(y) = 0` ...(i)
`N_(1) sin 30^(@) + N_(2) - mg =0` ...(ii)
Also tau_(0)=0`
`:.` mgxx(1)/(2)cos60^(@)-N_(1)((h)/(cos30^(@)))=0` ..(iii)
Also `N_(1) N_(2)`
From (i) `f = N_(1) cos 30^(@) = N_(1) sqrt3/(2)`
`N_(1) xx(1)/(2) + N_(1)= mg`
or `(3)/(2)N_(1) =mg`
From (iii) `(1)/(2) xx (1)/(2) = N_(1) (h.2)/(sqrt3) = (2)/(3)mg. (2h)/(sqrt30)`
`(l)/(4) =(4h)/(3sqrt3) :. (h)/(l) =(3sqrt3)/(16)`
We can show that `f =(16sqrt3)/(3)`
.