Viscous Flow Through A Pipe

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Water flows steadily through a horizontal capillary tube under a fixed pressure difference between its ends. If the tube is replaced by another of the same length but half the internal radius, by what factor does the volume flow rate change?

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Steady laminar flow of a viscous liquid through a narrow tube is governed by Poiseuille's equation, $Q = \frac{\pi P r^4}{8\eta L}$, where $P$ is the pressure difference, $r$ the radius, $L$ the length and $\eta$ the viscosity. The striking feature is the fourth-power dependence on radius, which arises because both the cross-sectional area and the velocity profile favour wider tubes. With pressure, length and viscosity held constant, $Q \propto r^4$. Replacing the tube with one of half the radius gives $Q_2/Q_1 = (1/2)^4 = 1/16$. The option 1/2 assumes flow scales with radius, ignoring area and velocity effects. The option 1/4 treats $Q$ as proportional to area alone. The option 1/8 corresponds to an $r^3$ law and is still too generous. This dramatic $r^4$ sensitivity is exactly the result emphasised in the NCERT discussion of viscous flow and is why narrowed arteries so severely restrict blood flow. The fourth power emerges because the cross-sectional area itself contributes an $r^2$ factor while the parabolic velocity profile contributes a further $r^2$, since fluid near the centre of a wider tube faces far less wall drag. As a plausibility check, halving the radius slashes the flow far more than halving it, and the factor 1/16 reflects this extreme sensitivity correctly.

Viscous Flow Through A Pipe

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