Two very large open tanks `A & F` both contain the same liquid. A horizontal pipe `BCD`, having a small constriction at `C`, leads out of the bottom of tank `A`, and a vertical pipe `E` containing air opens into the constriction at `C` and dips into the liquid in tank `F`. Assume stremline flow and no viscosity. If the cross section area at `C` is one-half that at `D`, and if `D` is at distance `h_(1)` below the level of the liquid in `A`, to what height `h_(2)` will liquid rise in pipe `E`? Express your answer in terms of `h_(1)`.
[Neglect changes in atmosphere pressure with elevation. In the containers there is atmosphere above the water surface and `D` is also open to atmosphere.]

Answer

`h_(2) = 3h_(1)`

Explanation

Apply Bernolli's equation `b//w` point `P` & `D`
`P_(atm) + rhogh_(1) + (1)/(2)rhoV_(P)^(2) = P_(atm) + rhog(o) + (1)/(2)rhov_(D)^(2)`
[Assume zero level `D`]
`V_(p) approx 0` Area of cross section is very large
`V_(D) = sqrt(2gh_(1)`
Since area at `C` is half than area at `D`
so according to continuity equation
`V_(C)A_(C) = V_(D)A_(D) rArr V_(C) = 2V_(D) = 2sqrt2gh_(1)`
Now for point `P` & `C` according to Bernolli's equation.
`P_(P) + rhogh_(1) + (1)/(2)rhoV_(p)^(2) = P_(C) + rhog(0) + (1)/(2)rho.V_(C)^(2)`
`V_(P) approx 0`
`rArr P_(atm) + rhogh_(1) = P_(C) + (1)/(2)rho.(2sqrt(2gh_(1)))^(2)`
`rArr P_(atm) = P_(C) + 3rhogh_(1) ....(i)`
for point `E`
`P_(E) = P_(C) + rhogh_(2) = P_(atm)`
from `(i)` & `(ii)`
`3rhogh_(1) = rhogh_(2)`
`rArr h_(2) = 3h_(1)`