Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h_{0} and pressure 2p_{0} where p_{0} is the atmospheric pressure. There is a hole in the wall of the tank at a depth h_{1} below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h_{2} of the water in the long tube above the top initially. (b) Find the speed with which water comes out of the hole. (c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.

#### Solution

(a) Pressure of water above the water level of the bigger tank is given by

`P = (h_2 + h_0)rhog`

Let the atmospheric pressure above the tube be `P_0`.

Total pressure above the tube = `P_0 + P`

=`(h_2+h_0)rhog+P_0`

This pressure initially is balanced by pressure above the tank `2P_0`.

⇒ `2P_0 = (h_2 + h_0)rhog + P_0`

⇒ `h_2 = (P_0)/(rhog) - h_0`

(b) Velocity of the efflux out of the outlet depends upon the total pressure above the outlet.Total pressure above the outlet = `2P_0 + (h_1 - h_0)rhog`

Applying Bernouli's law, we get

Let the velocity of efflux be v_{1} and the velocity with which the level of the tank falls be v_{2} . pressure above the outlet is `P_0`. Then,

`(2P_0+(h_1-h_0)rhog)/rho` + gz + `v_2^2/2` = `P_0/rho`+gz+`v_2^2/2`

Now , let the reference point of the liquid be the level of the outlet . Thus, z = 0

⇒ `(P_0+(h_1 - h_0)rhog)/(rho)` + `v_2^2/2` = `v_1^2/2`

Again , the speed with which the water level of the tank goes down is very less compared to the velocity of the efflux . Thus,

`"v"_2 = 0`

⇒ `(P_0+(h_1-h_0)rhog)/(rho)` = `v_1^2/2`

⇒ `v_1` = `[2/rho(P_0 + (h_1 - h_0) rhog)]^(1/2)`

(c) Water maintains its own level , so height of the water of the tank will be `"h"_1` when water will stop flowing

Thus height of water in the tube below the tank height will be = h_{1}

Hence height of the water above the tank height will be = -h_{1}