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Q.

A narrow tube of radius R and length L is connected in series with another tube of radius $\frac{R}{2}$ and length $\frac{L}{4}$ If the pressure across the two tubes taken together is P, the pressure across the two tubes separately are (Assume steady and stream line flow)

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a

$\frac{4}{5} P and \frac{P}{5}$

b

$\frac{P}{3} and \frac{2}{3} P$

c

$\frac{P}{4} and \frac{3}{4} P$

d

$\frac{P}{2} and \frac{P}{2}$

answer is A.

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Detailed Solution

$Given P_{1} - P_{3} = P$

Question Image

We know that $P_{1} - P_{3} = (P_{1} - P_{2}) + (P_{2} - P_{3})$

$\Rightarrow (P_{1} - P_{2}) + (P_{2} - P_{3}) = P . . . . . . . . (1)$

$From \frac{v}{t} = \frac{π \Pr^{4}}{8 η ℓ}$

$\frac{π (P_{1} - P_{2}) R^{4}}{8 η L} = \frac{π (P_{2} - P_{3}) {(\frac{R}{2})}^{4}}{8 η \frac{L}{4}}$

$\frac{π (P_{1} - P_{2}) R^{4}}{8 η L} = 4 \frac{π (P_{2} - P_{3}) R^{4}}{16 \times 8 η L}$

$\Rightarrow P_{1} - P_{2} = \frac{P_{2} - P_{3}}{4} . . . . . . . . . (2)$

From 1 & 2

$\Rightarrow P = (P_{1} - P_{2}) + 4 (P_{1} - P_{2}) \Rightarrow P_{1} - P_{2} = P / 5$

$and \frac{P}{5} = \frac{P_{2} - P_{3}}{4} \Rightarrow P_{2} - P_{3} = \frac{4}{5} P$

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