Question

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.


Find the height $h_2$ of the water in the long tube above the top initially.

• A. $\frac{3P_0}{\rho g}-\frac{h_0}{3}$
• B. $\frac{2P_0}{\rho g}-\frac{h_0}{2}$
• C. $\frac{P_0}{\rho g}-h_0$
• D. $\frac{P_0}{2\rho g}-2h_0$

Answer 1

The correct option is C $\frac{P_0}{\rho g}-h_0$


Form the above diagram we can see that, $P_A=P_B$
and $P_B=2P_0$ (Given)
If, $P_0$ is the atmoshperic pressure then $P_A$ will be
$P_A=P_0+\rho \times g\times (h_2+h_0)=2P_0$
$\rho \times g\times (h_2+h_0)=P_0$
$(h_2+h_0)=\frac{P_0}{\rho \times g}$
$h_2=\frac{P_0}{\rho \times g}-h_0$