Question

Water flows in a streamline motion through a horizontal pipe of circular cross-section as shown in the figure. The pressure difference of water between \(P\) and \(Q\) is \(15\,\text{N m}^{-2}\). The area of cross-section at \(P\) and \(Q\) are \(40\,\text{cm}^2\) and \(20\,\text{cm}^2\), respectively. The rate of flow of water through the pipe, in \(\text{cm}^3\text{s}^{-1}\), is:

• A. \(400\)
• B. \(100\)
• C. \(200\)
• D. \(300\)

Hint:

Use continuity before Bernoulli equation. Smaller area means greater speed. For horizontal pipes, height terms cancel. Convert units carefully.

Answer 1

The Correct Option is C

Concept:

• Apply Bernoulli's theorem.

• Use equation of continuity.

• For horizontal flow, gravitational term remains constant.

Step 1: Apply continuity equation
\[ A_1v_1=A_2v_2 \] \[ 40v_1=20v_2 \] \[ v_2=2v_1 \]

Step 2: Apply Bernoulli equation
\[ P_1+\frac12\rho v_1^2 = P_2+\frac12\rho v_2^2 \] \[ 15 = \frac12(1000) \left(v_2^2-v_1^2\right) \] \[ 15 = 500(4v_1^2-v_1^2) \] \[ 15=1500v_1^2 \] \[ v_1=0.1\,\text{m s}^{-1} \]

Step 3: Calculate discharge
\[ Q=A_1v_1 \] \[ =(40\times10^{-4})(0.1) \] \[ =4\times10^{-4}\,\text{m}^3\text{s}^{-1} \] \[ =400\,\text{cm}^3\text{s}^{-1} \]