Question

Water flows through a horizontal tube as shown in the figure. The difference in height between the water columns in vertical tubes is 5 cm and the area of cross-sections at A and B are 6 cm\(^2\) and 3 cm\(^2\) respectively. The rate of flow will be ______ cm\(^3\)/s. (take g = 10 m/s\(^2\)).

• A. \(200\sqrt{6}\)
• B. \(100\sqrt{3}\)
• C. \(200/\sqrt{3}\)
• D. \(200\sqrt{3}\)

Hint:

For venturi meter problems, it's useful to remember the derived formula for flow rate:
\[ Q = A_A A_B \sqrt{\frac{2gh}{A_A^2 - A_B^2}} \] This formula combines the continuity and Bernoulli equations and can save time in calculations. However, understanding the derivation from first principles is more important.

Answer 1

The Correct Option is D

The problem is solved using Bernoulli’s equation and the continuity equation.

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Given:

• Height difference between water columns: \[ h = 5~\text{cm} = 0.05~\text{m} \]
• Area at point A: \[ A_1 = 6~\text{cm}^2 = 6 \times 10^{-4}~\text{m}^2 \]
• Area at point B: \[ A_2 = 3~\text{cm}^2 = 3 \times 10^{-4}~\text{m}^2 \]
• Acceleration due to gravity: \[ g = 10~\text{m/s}^2 \]

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Principles Used:

1. Bernoulli’s equation for a horizontal tube: \[ P_1 + \frac{1}{2}\rho v_1^2 = P_2 + \frac{1}{2}\rho v_2^2 \]
2. Continuity equation: \[ A_1 v_1 = A_2 v_2 \]

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Step-by-Step Solution:

1. The pressure difference due to the height difference is: \[ \Delta P = \rho g h \]
2. From Bernoulli’s equation: \[ \rho g h = \frac{1}{2}\rho (v_2^2 - v_1^2) \]
3. Cancelling \( \rho \): \[ 2gh = v_2^2 - v_1^2 \]
4. From continuity: \[ v_1 = \frac{A_2}{A_1} v_2 = \frac{3}{6} v_2 = \frac{v_2}{2} \]
5. Substituting into the Bernoulli equation: \[ 2gh = v_2^2 - \left(\frac{v_2}{2}\right)^2 \]
6. Simplifying: \[ 2 \times 10 \times 0.05 = \frac{3v_2^2}{4} \] \[ 1 = \frac{3v_2^2}{4} \]
7. Solving for \( v_2 \): \[ v_2^2 = \frac{4}{3} \quad \Rightarrow \quad v_2 = \frac{2}{\sqrt{3}}~\text{m/s} \]
8. The volume flow rate is: \[ Q = A_2 v_2 \] \[ Q = 3 \times 10^{-4} \times \frac{2}{\sqrt{3}}~\text{m}^3/\text{s} \]
9. Converting to \( \text{cm}^3/\text{s} \): \[ Q = 200\sqrt{3}~\text{cm}^3/\text{s} \]

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Final Answer:

\(\boxed{200\sqrt{3}~\text{cm}^3/\text{s}}\)