Question:medium

A cylindrical pipe has a radius of \( 0.1 \, \text{m} \). If the speed of water flowing through the pipe is \( 2 \, \text{m/s} \), calculate the volume flow rate of water through the pipe.

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The volume flow rate is directly proportional to the cross-sectional area and the velocity of the fluid. Larger pipes or faster flowing fluids result in higher flow rates.
Updated On: Nov 26, 2025
  • \( 0.0628 \, \text{m}^3/\text{s} \)
  • \( 0.0314 \, \text{m}^3/\text{s} \)
  • \( 0.1256 \, \text{m}^3/\text{s} \)
  • \( 0.02 \, \text{m}^3/\text{s} \)
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The Correct Option is A

Solution and Explanation

The volume flow rate \( Q \) is calculated using the formula: \[ Q = A \times v \] where \( Q \) denotes volume flow rate, \( A \) represents the cross-sectional area of the pipe, and \( v = 2 \, \text{m/s} \) is the water speed. For a cylindrical pipe, the cross-sectional area \( A \) is determined by: \[ A = \pi r^2 \] with \( r = 0.1 \, \text{m} \) being the pipe's radius. Substituting the value of \( r \): \[ A = \pi \times (0.1)^2 = \pi \times 0.01 = 0.0314 \, \text{m}^2 \] The volume flow rate is then computed as: \[ Q = 0.0314 \times 2 = 0.0628 \, \text{m}^3/\text{s} \] Therefore, the volume flow rate of water through the pipe is \( 0.0628 \, \text{m}^3/\text{s} \).
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