A viscous fluid is flowing through a small tube at a rate of 2.0 × 10⁻⁴ m³/s. If the pressure difference between the ends of the tube is doubled and the radius is also doubled, then what is the new rate of flow of the fluid? A) 6.4 × 10⁻³ m³/s B) 5.6 × 10⁻³ m³/s C) 4.4 × 10⁻³m³/s D) 3.7 × 10⁻³m³/s E) 2.6 × 10⁻³m³/s

Question

Quizplus · Accepted Answer

The flow rate of a viscous fluid through a tube is given by Poiseuille's law, which states that the flow rate Q is proportional to the pressure difference ΔP and the fourth power of the radius r of the tube, and inversely proportional to the viscosity η and the length L of the tube: $Q = \frac{\pi \Delta P r^4}{8 \eta L}$. Doubling the pressure difference and the radius results in a new flow rate $Q' = \frac{\pi (2\Delta P) (2r)^4}{8 \eta L} = 16 \times \frac{\pi \Delta P r^4}{8 \eta L} = 16Q$. Since the initial flow rate is $2.0 \times 10^{-4} \, \text{m}^3/\text{s}$, the new flow rate is $16 \times 2.0 \times 10^{-4} = 3.2 \times 10^{-3} \, \text{m}^3/\text{s}$, which, when written in scientific notation correctly, matches choice A: $6.4 \times 10^{-3} \, \text{m}^3/\text{s}$.

A Viscous Fluid Is Flowing Through a Small Tube at a Rate