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Pressure at a certain depth in liquid - class-XII
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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
💡 Explanation:
Here, $l _1 = l _2 = 1m$ and $\displaystyle \frac{r _1}{r _2} = \frac{1}{2}$
As $V = \displaystyle \frac{ \pi P _1 r _1^4 }{8 \eta l} = \frac{ \pi P _2 r _2^4 }{8 \eta l}$ or $\displaystyle \frac{ P _1 }{P _2 } = \left( \frac{ r _2 }{ r _1} \right)^4 = 16$
$\therefore P _1 = 16 P _2$
Since, both tubes are connected in series, hence pressure difference across the combination is
$P = P _1 + P _2$ $\Rightarrow$ $\displaystyle 1 = P _1 + \frac{P _1}{16}$
or $\displaystyle P _1 = \frac{16}{17} = 0.94 m$