Question

Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h0 and pressure 2p0 where p0 is the atmospheric pressure. There is a hole in the wall of the tank at a depth h1 below the top from which water comes out. A long vertical tube is connected as shown. (a) Find the height h2 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.
Figure

Answer 1

(a) The height is $h_{2}=\frac{P_{0}}{\rho g}-h_{O}$

(b) The speed is $2 P_{0} \pm\left(h_{1}-h_{0}\right) \rho g$

(c) Height of the water in the long tube is  −$h_1$

Explanation:

(a) Pressure of water above the greater tank's water level is given by

$P=\left(h_{2}+h_{0}\right) \rho g$

Let the pressure atmospheric over the tube be$P_0$ .

Total pressure above the tube  = $P_0+P$

$\text { Total Pressure }=\left(\mathrm{h}_{2}+\mathrm{h}_{0}\right) \mathrm{Pg}+\mathrm{P}_{0}$

Initially this pressure is balanced by a pressure $2P_o$ above the  tank.

$2 P_{O}=\left(h_{2}+h_{O}\right) \rho g+P_{O}$

$h_{2}=\frac{P_{0}}{\rho g}-h_{O}$

(b) The speed of the efflux from the outlet depends on the total pressure above the outlet.

$=2 P_{0} \pm\left(h_{1}-h_{0}\right) \rho g$

By applying Bernouli's law, we get  

Let the efflux velocity be $v_1$ and the velocity at which the tank level falls to be $v_2$. Over the exit pressure is Po. Then

$\frac{2 P_{O}+\left(h_{1}-h_{O}\right) \rho g}{\rho}+g z+\frac{v_{2}^{2}}{2}=\frac{P_{O}}{\rho}+g z+\frac{v_{1}^{2}}{2}$

Now, let outlet level be the reference point for the liquid. So, z=0

$\frac{2 \mathrm{P}_{\mathrm{O}}+\left(\mathrm{h}_{1}-\mathrm{h}_{\mathrm{O}}\right) \rho \mathrm{g}}{\rho}+\frac{\mathrm{v}_{2}^{2}}{2}=\frac{\mathrm{v}_{1}^{2}}{2}$

Again, the speed at which the tank's water level falls is much lower compared with the efflux velocity. Accordingly,

$V_2 = 0$

$\frac{2 P_{O}+\left(h_{1}-h_{O}\right) \rho g}{\rho}=\frac{v_{1}^{2}}{2}$

$v_{1}=\sqrt{\frac{2}{\rho} 2 P_{O}+\left(h_{1}-h_{O}\right) \rho g}$

(c) Water keeps its own level , So the tank's water height will be $h_1$ when the water begins to flow

Then water height in the tube below tank height is = $h_1$

Therefore water height above tank height is = −$h_1$

Therefore the height of the water in the long tube above the top initially is  $h_{2}=\frac{P_{0}}{\rho g}-h_{O}$ and the speed with which water comes out of the hole is $2 P_{0} \pm\left(h_{1}-h_{0}\right) \rho g$  and the height of the water in the long tube above the top when the water stops coming out of the hole is −$h_1$