A viscous fluid is flowing through a small tube at a rate of 0.400 m³/s.If the pressure difference between one end of the tube and the other is tripled,the length is doubled and the radius is also doubled,then what is the rate of flow of the fluid? A) 9.6 m³/s B) 10.2 m³/s C) 12.00 m³/s D) 14.2 m³/s E) 15.5 m³/s

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Quizplus · Accepted Answer

The rate of flow of fluid is given by the Poiseuille's equation: $\Delta P=\frac{8\eta LQ}{\pi r^4}$ where $\Delta P$ is the pressure difference, $\eta$ is the viscosity of the fluid, $L$ is the length of the tube, $Q$ is the rate of flow of fluid, and $r$ is the radius of the tube. When the pressure difference is tripled, the new pressure difference becomes $3\Delta P$. The length and radius both are doubled, so the new length is $2L$ and the new radius is $2r$. Substituting these new values into the Poiseuille's equation, we get: $3\Delta P=\frac{8\eta(2L)Q'}{\pi (2r)^4}$ Simplifying, we get: $Q'=\frac{Q}{2^5}3=0.075Q$ Substituting the given value of $Q$ in the equation, we get: $Q'=0.075 imes0.400=0.03\ ext{m}^3/ ext{s}=9.6\ ext{mL/s}=9.6\ ext{mS}\upsilon$ Therefore, the rate of flow of fluid is 9.6 m³/s.

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