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Q.

Two capillaries of same length and radii in the ratio 1 : 2 are connected in series. A liquid flows through them in streamlined condition. If the pressure across the two extreme ends of the combination is 1 m of water, the pressure difference across first capillary is

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a

0.94 m

b

4.9 m

c

0.49 m

d

9.4 m

answer is D.

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Detailed Solution

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Given, $l_{1} = l_{2} = 1,$ and $\frac{r_{1}}{r_{2}} = \frac{1}{2}$
$V = \frac{π P_{1} r_{1}^{4}}{8 η l} = \frac{π P_{2} r_{2}^{4}}{8 η l} \Rightarrow \frac{P_{1}}{P_{2}} = {(\frac{r_{2}}{r_{1}})}^{4} = 16$
$\Rightarrow P_{1} = 16 P_{2}$
Since both tubes are connected in series, hence pressure difference across combination, ${P=P}_{1} + P_{2} \Rightarrow 1 = P_{1} + \frac{P_{1}}{16} \Rightarrow P_{1} = \frac{16}{17} = 0 .94 m$

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