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A capillary tube is attached horizontally to a constant pressure head arrangement. If the radius of the capillary tube is increased by $10\% $, then the rate of flow of the liquid shall change nearly by:
A) $ + 10\% $
B) $46\% $
C) $-10\% $
D) $-40\% $

Last updated date: 18th Jun 2024
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Hint: The rate of flow of liquid is directly proportional to the change in pressure at the required point. In other words, if you want to calculate rate of flow at a point then you must calculate the pressure difference or loss of pressure at that point of the flowing stream.

Complete step by step solution:
The equation in Poiseuille defines a pressure drop regardless of the viscosity of the fluid; other types of pressure drops which also exist in a fluid. The pressure needed to force a fluid viscous to gravity must, for example, include all those required by the Poiseuille Rule and those required by the Bernoulli team to provide pressure at each point in the flow.
Another example is that if a blood goes through a smaller constriction, its speed is higher (due to the consistency of the volumetric flow rate) than that of a wide diameter and its pressure is lower than that of a broader diameter (due to the equation of Bernoulli) However the blood viscosity induces additional pressure decreases in the flow direction, which is proportional to the duration of the flow. The real pressure reduction has been triggered by both effects.
Unless we take into account viscosity of the fluid a constant pressure head should be placed in a horizontal conduit. Poiseuille's equation can be stated also as $\Delta P = \dfrac{{8\mu LQ}} {{\pi {R^4}}} $
$\Delta P$ is the pressure loss,
$\mu $ is the dynamic viscosity
$Q$ is the volumetric flow rate,
$L$ is the length of pipe,
$R$ is the radius,
$\Delta P$, $\mu $, $L$ are constants,
Therefore, \[Q \propto \ ;{ R^4}\]
$ \Rightarrow Q = K{R^4} $
$ \Rightarrow$ \[R\prime = R + 0.1R = 1.1R\; \]
$ \Rightarrow$ \[Q\prime = K {(1.1R) ^4}\]
$ \Rightarrow$ $\dfrac {{Q' - Q}} {Q} = \dfrac {{{{(1.1R)} ^4} - {R^4}}} {{{R^4}}} $
$ \therefore$ $0.4641$
 The rate of flow of the liquid shall change nearly $46\% $.
Hence the correct option is B.

Note: The equation assumes that the fluid is incompressible and Newtonian; the flow into a constant loop cross-section pipe is laminar which is considerably longer than the diameter and that the fluid does not accelerate in the pipe. The present fluid stream is not laminar, but unstable in speeds and tube diameters above a threshold, resulting in greater pressure drops according to the Hagen-Poiseuille equation.