A fluid is flowing with an average speed of 1.5 m/s through a tube that has a radius of 2.0 mm and is 18 cm long.The drop in pressure is 967 Pa.What is the viscosity of the fluid? A)0.0018 N ∙ s/m² B)0.0056 N ∙ s/m² C)0.028 N ∙ s/m² D)0.013 N ∙ s/m² E)0.0034 N ∙ s/m²

Question

Quizplus · Accepted Answer

The viscosity of the fluid can be calculated using Poiseuille's law, which relates the flow rate, pressure drop, viscosity, length of the tube, and the radius of the tube. The formula is given by $\Delta P = \frac{8 \mu L Q}{\pi r^4}$, where $\Delta P$ is the pressure drop, $\mu$ is the viscosity, $L$ is the length of the tube, $Q$ is the volume flow rate, and $r$ is the radius of the tube. Rearranging the formula to solve for $\mu$ gives $\mu = \frac{\Delta P \pi r^4}{8 L Q}$.Given:- $\Delta P = 967 \, \text{Pa}$- $r = 2.0 \, \text{mm} = 2.0 \times 10^{-3} \, \text{m}$- $L = 18 \, \text{cm} = 0.18 \, \text{m}$- Average speed $v = 1.5 \, \text{m/s}$First, calculate the volume flow rate $Q = vA = v\pi r^2$, where $A$ is the cross-sectional area of the tube.$Q = 1.5 \times \pi \times (2.0 \times 10^{-3})^2 = 1.5 \times \pi \times 4 \times 10^{-6} = 6\pi \times 10^{-6} \, \text{m}^3/\text{s}$Now, substitute the values into the rearranged Poiseuille's law formula to find $\mu$:$\mu = \frac{967 \times \pi \times (2.0 \times 10^{-3})^4}{8 \times 0.18 \times 6\pi \times 10^{-6}}$Solving this gives $\mu \approx 0.0018 \, \text{N} \cdot \text{s/m}^2$, which corresponds to choice A.

A Fluid Is Flowing with an Average Speed of 1