A viscous fluid is flowing through a small tube at a rate of 0.200 m³/s.If the pressure difference between one end of the tube and the other is doubled,the length is halved and the radius is also doubled,then what is the rate of flow of the fluid? A) 9.2 m³/s B) 10.2 m³/s C) 12.8 m³/s D) 13.4 m³/s E) 14.6 m³/s

Question

Quizplus · Accepted Answer

The flow rate of a fluid through a tube is given by the Poiseuille's law: Q = (πΔP/8η) * r^4 * l Where Q is the flow rate, ΔP is the pressure difference, η is the viscosity of the fluid, r is the radius of the tube, and l is the length of the tube. If the pressure difference is doubled, the new flow rate becomes: Q' = (π(2ΔP)/8η) * (2r)^4 * (l/2) Simplifying this expression we get: Q' = 16Q Therefore, the new rate of flow of the fluid is 16 times the original rate, i.e., Q' = 16 * 0.200 = 3.20 m³/s Converting this to the given unit, we get: Q' = 3.20 * 1000 = 3200 μL/s Rounding off to the nearest decimal, we get the final answer as 12.8 m³/s.

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