Question:

Water is flowing through a horizontal pipe in streamline flow. At the narrowest part of the pipe:

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In streamline flow, an increase in velocity corresponds to a decrease in pressure as per Bernoulli's principle. This is especially evident in areas with a smaller cross-section.
Updated On: Apr 14, 2025
  • Velocity is maximum and pressure is minimum.
  • Pressure is maximum and velocity is minimum.
  • Both pressure and velocity are minimum.
  • Both pressure and velocity are maximum.
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The Correct Option is A

Solution and Explanation

Step 1: Understanding Bernoulli's principle. 
Bernoulli's principle states that for an incompressible, non-viscous fluid in streamline flow: \[ P + \frac{1}{2} \rho v^2 + \rho gh = \text{constant}, \] where: \(P\): Pressure of the fluid, \(\rho\): Density of the fluid, \(v\): Velocity of the fluid, \(h\): Height of the fluid. 

Step 2: Application to a horizontal pipe. 
For a horizontal pipe (\(h\) is constant), the equation simplifies to: \[ P + \frac{1}{2} \rho v^2 = \text{constant}. \] This implies that when the velocity (\(v\)) of the fluid increases, the pressure (\(P\)) decreases, and vice versa. 

Step 3: Narrowest part of the pipe. 
At the narrowest part of the pipe: The cross-sectional area is minimum. By the equation of continuity (\(A_1v_1 = A_2v_2\)), the velocity is maximum. By Bernoulli's principle, the pressure is minimum. 

Step 4: Conclusion. 
At the narrowest part of the pipe, the velocity is maximum, and the pressure is minimum. \[ \therefore \text{The correct answer is: (A)}. \]

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