A tube of uniform diameter D is immersed in a steady flowing inviscid liquid stream of velocity V, as shown in the figure. Gravitational acceleration is represented by 𝑔. The volume flow rate through the tube is ______.

The correct option is:
$\frac{\pi }{4}{D}^2\sqrt{V^2-2g{h}_2}$

Step-by-Step Derivation and Explanation:

1. Identify the reference points and parameters from the diagram:
Let us analyze the system shown in the figures:
- The liquid stream flows horizontally with a steady velocity $V$.
- The free surface of the liquid stream is exposed to the atmosphere, where the pressure is $P_{\mathrm{atm}}$.
- Point 1 is chosen at the inlet of the submerged tube, which is at a depth of $h_1$ below the free surface.
- Point 2 is chosen at the exit of the tube, which is open to the atmosphere and situated at a height of $h_2$ above the free surface.
- The tube has a uniform diameter $D$. Thus, by the continuity equation, the velocity of the fluid inside the tube remains constant throughout, meaning $V_1=V_2$.

2. Bernoulli's Equation between the upstream stream and the tube inlet (Point 1):
Consider an upstream point at the same elevation as the tube inlet (depth $h_1$). The pressure at this upstream point is hydrostatic:
$P_0=P_{\mathrm{atm}}+\rho g{h}_1$
The flow velocity at the upstream point is $V$. Applying Bernoulli's equation along the streamline from the upstream flow to the tube inlet (Point 1):
$\frac{P_0}{\rho g}+\frac{V^2}{2g}=\frac{P_1}{\rho g}+\frac{{V_1}^2}{2g}$
Substituting $P_0$ into the equation:
$\frac{P_{\mathrm{atm}}}{\rho g}+h_1+\frac{V^2}{2g}=\frac{P_1}{\rho g}+\frac{{V_1}^2}{\mathrm{2<!--\; <mi-->g}}$ ---- (Equation 1)

3. Bernoulli's Equation along the tube (Point 1 to Point 2):
Now, apply Bernoulli's equation inside the tube from the inlet (Point 1) to the exit (Point 2):

Since the tube has a uniform cross-section, the velocity inside is constant, so $V_1=V_2$. The exit at Point 2 is open to the atmosphere, so $P_2=P_{\mathrm{atm}}$. The elevation difference is $Z_2-Z_1=h_1+h_2$.
Substituting these values, the equation simplifies to:
$\frac{P_1}{\rho g}=\frac{P_{\mathrm{atm}}}{\rho g}+(h_1+h_2)$ ---- (Equation 2)

4. Solve for flow velocity inside the tube:
Substitute the expression for $\frac{P_1}{\rho g}$ from Equation 2 into Equation 1:

Subtracting $\frac{P_{\mathrm{atm}}}{\rho g}+h_1$ from both sides gives:
$\frac{V^2}{\mathrm{2<!--\; <mi-->g}}=h_2+\frac{{V_1}^2}{\mathrm{2<!--\; <mi-->g}}$
Rearranging to solve for the velocity inside the tube $V_1$:

Multiplying by $2g$ and taking the square root:

5. Calculate the volume flow rate:
The volume flow rate $Q$ is the product of the tube's cross-sectional area and the fluid velocity inside:
$Q=A·V_1$
Since the cross-sectional area for a tube of uniform diameter $D$ is $A=\frac{\pi }{4}{D}^2$: