Find volume flow rate in the venturi meter given below in which water is flowing.

[Cross sectional area at A & B is A and a, A/a = 2.4A = √3 m². ρ = 1000 kg/m³]

Correct Answer: Option 1 (1 m³/s)

To find the volume flow rate in the Venturi meter, we can apply Bernoulli's equation and the equation of continuity.

1. Identify the given parameters from the problem statement and diagram:
- Height difference between the water columns in the vertical tubes:
$h=5\text{cm}=0.05\text{m}$
- Cross-sectional area ratio:
$\frac{A}{a}=2$
- Cross-sectional area at point A:
$A=\sqrt{3}{\text{m}}^2$
- Density of water:
$\rho =1000{\text{kg/m}}^3$
- Acceleration due to gravity:
$g=10{\text{m/s}}^2$

2. Equation of Continuity:
The volume flow rate $Q$ remains constant at both points A and B:
$Q=A{v}_A=a{v}_B$
Since $\frac{A}{a}=2$, we have:
$v_B=2v_A$

3. Bernoulli's Equation:
Applying Bernoulli's principle along the streamline between point A (wider section) and point B (throat section):
$P_A+\frac{1}{2}\rho v_A^2=P_B+\frac{1}{2}\rho v_B^2$
Rearranging the terms:
$P_A-P_B=\frac{1}{2}\rho \left(v_B^2-v_A^2\right)$
The pressure difference is related to the height difference $h$ of the water column:
$P_A-P_B=\rho gh$
Equating the two expressions:
$\rho gh=\frac{1}{2}\rho \left(v_B^2-v_A^2\right)$
$2gh=v_B^2-v_A^2$

4. Solving for Velocity:
Substitute $v_B=2v_A$ into the equation:
$2gh={\left(2v_A\right)}^2-v_A^2$
$2gh=3v_A^2$
$v_A=\sqrt{\frac{2gh}{3}}$

5. Calculate Volume Flow Rate (Q):
Substitute the values into the equation for $Q$:
$Q=A\sqrt{\frac{2gh}{3}}$
$Q=\sqrt{3}\times \sqrt{\frac{2\times 10\times 0.05}{3}}$
$Q=\sqrt{3}\times \sqrt{\frac{1}{3}}$
$Q=\sqrt{3}\times \frac{1}{\sqrt{3}}=1{\text{m}}^3\text{/s}$