Question:

A pipe with diameter 25 cm at the entrance carries oil with specific gravity 0.9 at a velocity of 3 m/s. At the exit section the diameter of the pipe is 20 cm. The velocity at the exit section is:

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For incompressible flow, use the continuity equation to relate the velocity and cross-sectional areas at different points.
Updated On: Sep 24, 2025
  • 4.5 m/s
  • 4.68 m/s
  • 4.8 m/s
  • 4.9 m/s
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The Correct Option is B

Solution and Explanation


Step 1: Use the continuity equation.
The continuity equation for an incompressible flow is given by: \[ A_1 v_1 = A_2 v_2 \] Where: - \( A_1 = \frac{\pi d_1^2}{4} \) is the cross-sectional area at the entrance, - \( A_2 = \frac{\pi d_2^2}{4} \) is the cross-sectional area at the exit, - \( v_1 = 3 \, \text{m/s} \) is the velocity at the entrance, - \( d_1 = 0.25 \, \text{m} \) and \( d_2 = 0.20 \, \text{m} \) are the diameters at the entrance and exit, respectively.

Step 2: Solve for \( v_2 \).
Rearranging the continuity equation to solve for \( v_2 \): \[ v_2 = v_1 \times \frac{A_1}{A_2} = 3 \times \frac{\left( \frac{\pi (0.25)^2}{4} \right)}{\left( \frac{\pi (0.20)^2}{4} \right)} = 3 \times \left(\frac{0.25^2}{0.20^2}\right) = 3 \times \left(\frac{0.0625}{0.04}\right) = 4.68 \, \text{m/s} \]

Final Answer: \[ \boxed{4.68 \, \text{m/s}} \]

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