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Q.

Water flows in a streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to a/2 and the pressure increased to 2P, the rate of flow becomes

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a

$\frac{Q}{4}$

b

$Q$

c

$4 Q$

d

$\frac{Q}{8}$

answer is D.

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Detailed Solution

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$\begin{array}{l} V = \frac{{πpr}^{4}}{8 ηl} ∴ V \propto \Pr^{4} (η and l are constants) \\ ∴ \frac{V_{2}}{V_{1}} = (\frac{P_{2}}{P_{1}}) {(\frac{r_{2}}{r_{1}})}^{4} = 2 \times {(\frac{1}{2})}^{4} = \frac{1}{8} \\ ∴ V_{2} = \frac{Q}{8} \end{array}$

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