Poiseuille’s Law and Capillary Flow – Rankers Physics
Topic: Solid and Fluids
Subtopic: Fluid Dynamics

Poiseuille’s Law and Capillary Flow

A liquid is flowing in a horizontal uniform capillary tube under a constant pressure difference P. The value of pressure for which the rate of flow of the liquid is doubled when the radius and length both are doubled is:
P
\(\frac{3P}{4}\)
\(\frac{P}{2}\)
\(\frac{P}{4}\)

Solution:

Rate of flow \(Q = \frac{\pi P r^4}{8 \eta l}\). Under new conditions, \(Q' = 2Q\), \(r' = 2r\), and \(l' = 2l\), giving \(Q' = \frac{\pi P' (2r)^4}{8 \eta (2l)} = 8 \left(\frac{\pi P' r^4}{8 \eta l}\). Thus, \(8 P' = 2 P\), yielding \(P' = P/4\).

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