Question:medium

A pipe has a radius of 2 cm at one end and 1 cm at the other end. The velocity of the water at the wider end is 5 m/s. What is the velocity of the water at the narrower end, assuming incompressible flow?

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In incompressible flow, the product of cross-sectional area and velocity at any point in the pipe remains constant. Use the principle of continuity to relate velocities at different points in the pipe.
Updated On: Nov 26, 2025
  • \( 10 \, \text{m/s} \)
  • \( 20 \, \text{m/s} \)
  • \( 15 \, \text{m/s} \)
  • \( 25 \, \text{m/s} \)
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The Correct Option is A

Solution and Explanation

For incompressible fluids, the principle of continuity dictates that the flow rate (volume per unit time) is constant, expressed as: \[A_1 v_1 = A_2 v_2\] Where: - \( A_1 = \pi r_1^2 \) represents the cross-sectional area at the wider end. - \( A_2 = \pi r_2^2 \) represents the cross-sectional area at the narrower end. - \( v_1 = 5 \, \text{m/s} \) is the velocity at the wider end. - \( v_2 \) is the velocity at the narrower end. Substituting the given values: \[A_1 v_1 = A_2 v_2\] \[\pi (2)^2 \times 5 = \pi (1)^2 \times v_2\] \[4 \times 5 = 1 \times v_2\] \[v_2 = 20 \, \text{m/s}\] Therefore, the velocity of the water at the narrower end is \( 20 \, \text{m/s} \).
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