Question:

Water flows through a horizontal pipe with a velocity of 2 m/s. The cross-sectional area of the pipe reduces to half. What is the velocity of water in the narrower section?

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Use the continuity equation $A_1v_1 = A_2v_2$ to relate flow speeds in pipes of varying cross-section. If area halves, velocity doubles.
Updated On: June 02, 2025
  • 1 m/s
  • 2 m/s
  • 4 m/s
  • 8 m/s
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The Correct Option is C

Approach Solution - 1

According to the principle of conservation of mass (continuity equation) for an incompressible fluid: \[ A_1 v_1 = A_2 v_2 \] Given: \[ v_1 = 2 \, \text{m/s}, \quad A_2 = \frac{1}{2}A_1 \] Substitute into the equation: \[ A_1 \cdot 2 = \frac{1}{2}A_1 \cdot v_2 \quad \Rightarrow \quad 2 = \frac{1}{2} v_2 \quad \Rightarrow \quad v_2 = 4 \, \text{m/s} \]
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Approach Solution -2

To solve the problem, we need to find the velocity of water in the narrower section of a horizontal pipe when the cross-sectional area reduces to half.

1. Principle of Continuity: 
According to the continuity equation for incompressible fluids: 
$ A_1 v_1 = A_2 v_2 $
where $A$ is the cross-sectional area and $v$ is the velocity of the fluid.

2. Given Data:
Initial velocity, $v_1 = 2\, m/s$
Cross-sectional area reduces to half, so $A_2 = \frac{1}{2} A_1$

3. Calculate Velocity in Narrower Section:
Using continuity:
$ A_1 v_1 = A_2 v_2 \Rightarrow A_1 \times 2 = \frac{1}{2} A_1 \times v_2 $
Simplify:
$ 2 = \frac{1}{2} v_2 \Rightarrow v_2 = 4\, m/s $

Final Answer:
The velocity of water in the narrower section is $ {4\, m/s} $.

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