First slide

Viscous flow/Laminar flow

Question

A liquid flows steadily through a series combination of three capillary tubes of radii r, 2r and 3r, all of the same length L. If the pressure difference across the combination is 28 cm of mercury, the pressure difference (in cm of Hg) across the tube of radius 2r is very nearly equal to____cm of Hg

Moderate

Solution

$Q_{1} = Q_{2} = Q_{3}$ . Therefore,
$\begin{matrix} \frac{{πp}_{1} r^{4}}{8 ηL} = \frac{{πp}_{2} (2 r)^{4}}{8 ηL} = \frac{{πp}_{3} (3 r)^{4}}{8 ηL} \\ \Rightarrow p_{1} = 16 p_{2} = 81 p_{3} \\ ∴ p_{1} = 16 p_{2} \\ and p_{3} = \frac{16}{81} p_{2} \end{matrix}$
Given $p_{1} + p_{2} + p_{3} = 28 cm of Hg$
$\begin{array}{l} \Rightarrow 16 p_{2} + p_{2} + \frac{16}{81} p_{2} = 28 \\ \Rightarrow (16 + 1 + \frac{16}{81}) p_{2} = 28 \\ \Rightarrow p_{2} = \frac{81 \times 28}{1393} = 1 .628 cm of Hg \end{array}$

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