Question

Question: A pipe of length L = 25. 0 m that is open at one end contains air at atmospheric pressure. It is thrust vertically into a freshwater lake until the water rises halfway up in the pipe (Fig.). What is the depth h of the lower end of the pipe? Assume that the temperature is the same everywhere and does not change.

Short answer

Answer

The lower end of the pipe lies at the depth of h = 22.8 m.

Step-by-step solution

Concept

From figure 19-30, by using the formula for the balance of pressure inside and outside the pipe, and the condition for thepressure of the air inside the pipe, we can find thedepthof the lower end of the pipe.

1. The balance of the pressure inside and outside the pipe is

$\mathbf{p}\mathbf{+}{\mathbf{\rho }}_{\mathbf{w}}\mathbf{g}\left(\frac{\mathbf{L}}{\mathbf{2}}\right)\mathbf{=}{\mathbf{p}}_{\mathbf{0}}\mathbf{+}{\mathbf{\rho }}_{\mathbf{w}}\mathbf{gh}$

1. The pressure Pof the air inside the pipe satisfies

$\mathbf{p}\left(\frac{\mathbf{L}}{\mathbf{2}}\right)\mathbf{=}{\mathbf{p}}_{\mathbf{0}}\mathbf{L}$

Here,p is the pressure of air column inside the pipe ,p_w is the density of water, g is the acceleration due to gravity and L is the length of the pipe. Also, h is the depth of the water column outside the pipe.

Step 2: Given Data

1. The length of the pipe is L = 25.0 m
2. The pipe is open at one end, and it contains air at atmospheric pressure.
3. The water rises halfway in the pipe.
4. The temperature is the same everywhere and does not change.

Step 3: Calculations

We think about the open end of the pipe. The balance of the pressure inside and outside the pipe requires that

$\text{p}+{\rho }_w\text{g}\left(\frac{\text{L}}{2}\right)={\text{p}}_0+{\rho }_w\text{gh} ······································\left(1\right)$

Where p₀ is the atmospheric pressure, p is the pressure of air inside the pipe, L is the length of the pipe, and h is the depth of the lower end of the pipe.

The pressure p of air inside the pipe must satisfy the relation-

$p\left(\frac{L}{2}\right)=p_{0}L$

Hence,

$p=2p_0······································\left(2\right)$

Therefore, from Eq. (1) the depth of the lower end of the pipe is given by

$h=\frac{\left(p-p_0\right)}{{\rho }_{w}g}+\left(\frac{L}{2}\right)······································\left(3\right)$

By putting Eq. (2) in (3), we get

$$\begin{gathered} h=\frac{\left(2p_0-p_0\right)}{{\rho }_{w}g}+\left(\frac{L}{2}\right) \\ =\frac{p_0}{{\rho }_{w}g}+\left(\frac{L}{2}\right) \end{gathered}$$

For the given values the above equation becomes-

$$\begin{gathered} h=\frac{1.01\times {10}^5\text{Pa}}{1.00\times {10}^3\text{kg}/{\text{m}}^3\times 9.8\text{m}/\text{s}}+\left(\frac{25.0\text{m}}{2}\right) \\ =22.8\text{m} \end{gathered}$$

Conclusion

The lower end of the pipe lies at the depth of 22.8 m.