JEE Main 2020 · 6 September · evening · Single correct · +4 / −1

A fluid is flowing through a horizontal pipe of varying cross-section, with
speed v ms –1 at a point where the pressure is P pascal. At another point where pressure is P/2 Pascal its speed is V ms –1 . If the density of the fluid is ρ kg m –3 and the flow is streamline, then V is equal to :

  1. (a) √(P/2ρ + v²)
  2. (b) √(P/ρ + v²)
  3. (c) √(2P/ρ + v²)
  4. (d) √(P/ρ + v)
Answer and solution

Answer (b) √(P/ρ + v²)

Solution

Physics sir · 4 steps

Same height, so Bernoulli is a straight trade: the pressure you lose becomes kinetic energy per volume.

  1. Bernoulli, horizontal pipe. Both points are at the same height, so the ρ g h terms cancel. Pressure plus kinetic energy per unit volume stays the same.

    P + 1/2ρ v² = P/2 + 1/2ρ V²

  2. Pressure drop. Move the pressures to one side: 1/2ρ(V² - v²) = P - P/2 = P/2.
  3. Solve for V. Multiply both sides by 2/ρ to get V² - v² = P/ρ, so V² = P/ρ + v².
  4. So. V = √(P/ρ + v²), which is option (b).

The idea to keep. In a horizontal pipe, a pressure drop Δ P raises v² by (2Δ P)/ρ. Always use the pressure difference, and keep the 1/2 in 1/2ρ v².

Answer from the NTA final answer key. Solution by PrepWiser.

Study Fluid Mechanics

Asked about 90 times in JEE Main since 2020. This question is from 2020.