Correct Answer - B
Ans.(2)
At any instant of time let velocity of part of chain which is just coming in contact with surface `=v` and `dx=` length of chain reaches on surface in time `dt` then momentum carried by this part dp`=(dm)v`
`:.` rate of change of momentum of chain
`(dp)/(dt)=(dm)/(dt)v=(lamdadx)/(dt)v=lamdav=v=((m)/(l))v^(2)`
as `v=sqrt(2g((l)/(2)))sqrt(gl)` so `(dp)/(dt)=(m)/(l)(gl)=mg`
weight of part on the surface `=(mg)/(2)`
`:.` net force by chain on surface `=mg+(mg)/(2)=(3)/(2)mg`