MDCAT Physics MCQ #1861

A liquid flows through a horizontal tube of length 1 m and radius 0.005 m at a rate of 0.01 m³/s. What is the pressure difference?

MDCAT Physics MCQ #1861

  1. Question 1

    Q1. A liquid flows through a horizontal tube of length 1 m and radius 0.005 m at a rate of 0.01 m³/s. What is the pressure difference?

    • A) 1.02 x 10^4 Pa
    • B) 2.04 x 10^4 Pa
    • C) 4.08 x 10^4 Pa
    • D) 8.16 x 10^4 Pa

    Answer: 2.04 x 10^4 Pa

    Explanation: ΔP = (8ηLQ)/(πr⁴) = (8 * 0.001 Pa·s * 1 m * 0.01 m³/s) / (π * (0.005 m)⁴) = 2.04 x 10^4 Pa; correct application of Poiseuille's law.