A cylindrical vessel of a very large cross sectional area is containing two immiscible liquids of density `rho_(1)=600kg//m^(3)` and `rho_(2)=1200kg//m^(3)` as shown in the figure.
A small hole having cross sectional area `5cm^(2)` is made in right side vertical wall as shown. Take atmospheric pressure as `p_(0)=10^(5)N//m^(2), g=10m//s^(2)`. For this situation mark out the correct statement (s). Take cross sectional area of the cylindrical vessel as `1000cm^(2)`. Neglect the mass of the vessel.

Options

A. If the surface on which the vessel is placed is smooth, then a rightward force of magnitude `3 N` is to be applied on the vessel to maintain its static equilibrium.
B. If the surface on which the vessel is placed is smooth, then no force is needed to maintain its static equilibrium.
C. If the surface on which the vessel is placed is rough `(mu= 0.04)`, then the minimum force (horizontal) needed to be applied on the vessel to maintain its static equilibrium is zero.
D. If the surface on which the vessel is placed is rough `(=0.04)` then the maximum force (horizontal) needed to be applied on the vessel to maintain its static equilibrium is `19.8 N`.

Answer

A, C, D

Explanation

The velocity with which the liquid comes out is given by the expression
`P_(0)+rho_(1)gxx30cm=rho_(2)gxx10cm=p_(0)+(rho_(2)v^(2))/2`
`v=2.24m//s`
Due to transfer of liquid, change in moment takes place and hence a force has been exerted on the liquid in backward direction, this force is given by `F=rho_(2)av^(2)`, where a is the cross sectional area of hole `=1200xx5xx10^(-4)xx5=3N`
Normal contact force betwen the vessel and the suface is
`N=mg`
`N=(rho_(1)Axx0.3)g+(rho_(2)xxAxx0.2)G=420`
Where `A` is the cross sectional area of te vessel. Limiting friction force,
`f_(L)=muN=0.04xx420=16.8`
As `f_(L)gtF`, therefore the minimum force necessary to maintain the static equilibrium is `0`. For the maximum value of the applied force
`F_("max")=f_(L)+F=16.8+3=19.8n`