A viscous fluid is flowing through a small tube at a rate of 2.00 $\times$ 10^-4 m³/s.If the pressure difference between one end of the tube and the other is doubled and the radius is also doubled,then what is the rate of flow of the fluid? A) 64 $\times$ 10^-4 m³/s B) 56 $\times$ 10^-4 m³/s C) 44 $\times$ 10^-4 m³/s D) 37 $\times$ 10^-4 m³/s E) 26 $\times$ 10^-4 m³/s

Question

Quizplus · Accepted Answer

The rate of flow of a fluid through a tube can be given by the equation:
Q = πr^4ΔP/8μl
where Q is the rate of flow, r is the radius of the tube, ΔP is the pressure difference between the two ends of the tube, μ is the viscosity of the fluid, and l is the length of the tube.

If the pressure difference is doubled and the radius is doubled, then the new rate of flow can be found by plugging in the new values into the equation:
Q' = π(2r)^4(2ΔP)/8μl
Simplifying this expression gives:
Q' = 16Q
Substituting in the given value of Q, we get:
Q' = 16(2.00 x 10^-4 m^3/s) = 3.20 x 10^-3 m^3/s
Converting to scientific notation to match the answer choices, we get:
Q' = 32 x 10^-4 m^3/s = 32 x 10^-10 m^3/s = 64 x 10^-11 m^3/s
Therefore, the best choice is A: 64 x 10^-11 m^3/s.

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