Figure. shows a large closed cylindrical tank containing water. I… - Vidyadip

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.
Find the speed with which water comes out of the hole.
Find the height of the water in the long tube above the top when the water stops coming out of the hole.

✓

Answer

$2\text{P}_0\text{x} = (\text{h}_2 + \text{h}_0)\text{fg}$ [$\therefore$ Since liquid at the same level have same pressure]

$\Rightarrow2\text{P}_0 = \text{h}_2\text{fg} + \text{h}_0\text{fg}$
$\Rightarrow \text{h}_2\text{fg} = 2\text{P}_0 - \text{h}_0\text{fg}$
$\text{h}_2=\frac{2\text{P}_0}{\text{fg}}-\frac{\text{h}_0\text{fg}}{\text{fg}}=\frac{2\text{P}_0}{\text{fg}}-\text{h}_0$

K.E. of the water = Pressure energy of the water at that layer

$\Rightarrow\frac{1}{2}\text{mV}^2=\text{m}\times\frac{\text{P}}{\text{f}}$
$\Rightarrow\text{V}^2=\frac{2\text{P}}{\text{f}}=\Big[\frac{2}{\text{f}(\text{P}_0+\text{fg})(\text{h}_1-\text{h}_0)}\Big]$
$\Rightarrow\text{V}=\Big[\frac{2}{\text{f}(\text{P}_0+\text{fg})(\text{h}_1-\text{h}_0)}\Big]^{\frac{1}{2}}$

$(\text{x} + \text{P}_0)\text{fh} = 2\text{P}_0$

$\therefore2\text{P}_0 + \text{fg (h} - \text{h}_0)= \text{P}_0 + \text{fgx}$
$\therefore\text{X}=\frac{\text{P}_0}{\text{fg}+\text{h}_1-\text{h}_0}=\text{h}_2+\text{h}_1$
$\therefore$ i.e. x is $h_1$ meter below the top
⇒ x is $-h_1$ above the top

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