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 discharge rate from the outlet is

Options

A. `4.02xx10^(-3)m^(3)//s`
B. `6xx10^(-5)m^(3)//s`
C. `7.56xx10^(-3)m^(3)//s`
D. `4.02xx10^(-5)m^(3)//s`

Answer

A

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`