A liquid having density `6000 kg//m^(3)` stands to a height of `4 m` in a sealed tank as shown in the figure. The tank contains compressed air at a gauge pressure of `3 atm`. The horizontal outlet pipe has a cross-sectional area of `6 cm^(2)` and `3 cm^(2)` at larger and smaller sections. Atmospheric pressure `= 1 atm, g = 10m//s^(2) 1 atm =10^(5) N//m^(2)`.

Assume that depth of water in the tank remains constant due to `s` very large base and air pressure above it remains constant. Based on the above information, answer the following questions.
The height at which liquid will stand in the open end of the pipe is

Options

A. `4m`
B. `3.5m`
C. `4.5m`
D. `5.6m`

Answer

C

Explanation

Let atmospheric pressure be `p_(0)`. Then absolute pressure of air is `p+p_(0)=4atm`

Pressure at `A, p_(A)=p+p_(0)+rhogh`
From equation of continuity and Bernoulli's theorem,
`p_(B)=p_(C)`
Applying Bernoull's theorem at `A` and `E`, we have
`p+p_(0)+rhogh=p_(c)+(pv_(C)^(2))/2=p_(0)+(pv_(E)^(2))/2`
where `v_(C)` and `v_(E)` are the velocities of flow of liquid at `C` and `E`, respectively.
From hydrostastics `p_(c)=p_(0)+rhogh'`
From continuity equation `A_(C)v_(C)=A_(E)v_(E)`
Where `A_(C)=6cm^(2)` and `A_(E)=3cm^(2)`
Solving above equation, we get `v_(E)=13.4m//s`
`v_(C)=6.7m//s`
`h'=4.5m`
Discharge rate
`Q=A_(E)v_(E)=3xx10^(-4)xx13.4m^(3)//s`
When a hole has been created at the top of the vessel, gange pressure drops to zero, i.e,, in above equation put `p=0` and get `h'`.
`:. h'=3m`