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

Step 1: Understanding the Question:
The problem describes a venturi meter, a device used to measure the flow rate of a fluid. We are given the areas of the wide and narrow sections, and the pressure difference (indicated by the height difference in the vertical tubes). We need to find the volume flow rate.
Step 2: Key Formula or Approach:
We will use two fundamental principles of fluid dynamics: 1. Equation of Continuity: \( A_A v_A = A_B v_B = Q \), where Q is the volume flow rate. 2. Bernoulli's Equation for a horizontal tube: \( P_A + \frac{1}{2}\rho v_A^2 = P_B + \frac{1}{2}\rho v_B^2 \). The pressure difference is related to the height difference `h` by \( P_A - P_B = \rho g h \).
Step 3: Detailed Explanation:
Given values (in cgs units):
\( A_A = 6 \) cm\(^2\)
\( A_B = 3 \) cm\(^2\)
\( h = 5 \) cm
\( g = 10 \) m/s\(^2\) = 1000 cm/s\(^2\)
Density of water \( \rho = 1 \) g/cm\(^3\)
1. Relate velocities using the Equation of Continuity:
\[ 6 v_A = 3 v_B \implies v_B = 2 v_A \] 2. Use Bernoulli's Equation to find velocity:
Rearranging Bernoulli's equation:
\[ P_A - P_B = \frac{1}{2}\rho (v_B^2 - v_A^2) \] Substitute \( P_A - P_B = \rho g h \):
\[ \rho g h = \frac{1}{2}\rho (v_B^2 - v_A^2) \] \[ g h = \frac{1}{2} (v_B^2 - v_A^2) \] Now substitute \( v_B = 2v_A \):
\[ g h = \frac{1}{2} ((2v_A)^2 - v_A^2) = \frac{1}{2} (4v_A^2 - v_A^2) = \frac{3}{2} v_A^2 \] Solve for \( v_A \):
\[ v_A^2 = \frac{2gh}{3} \] \[ v_A = \sqrt{\frac{2gh}{3}} = \sqrt{\frac{2 \times 1000 \times 5}{3}} = \sqrt{\frac{10000}{3}} = \frac{100}{\sqrt{3}} \, \text{cm/s} \] 3. Calculate the Rate of Flow (Q):
\[ Q = A_A v_A = 6 \, \text{cm}^2 \times \frac{100}{\sqrt{3}} \, \frac{\text{cm}}{\text{s}} = \frac{600}{\sqrt{3}} \, \frac{\text{cm}^3}{\text{s}} \] To rationalize the denominator:
\[ Q = \frac{600 \sqrt{3}}{3} = 200\sqrt{3} \, \text{cm}^3/\text{s} \] Step 4: Final Answer:
The rate of flow is \( 200\sqrt{3} \) cm\(^3\)/s. This corresponds to option (D).