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Standard IX

Mathematics

Volume of a right circular cone

We have two n...

Question

We have two (narrow) capillary tubes $T_{1}$ and $T_{2}$ . their lengths are $l_{1}$ and $l_{2}$ and radii of cross section are $r_{1}$ and $r_{2}$ respectively. The rate of flow of water under a pressure difference $P$ through tube $T_{1}$ is $8 {c m}^{3} / s e c$ . If $l_{1} = 2 l_{2}$ and $r_{1} = 2 r_{2}$ . What will be the rate of flow when the two tuves are connected in series and pressure difference across the combinatin is same as before $[= P]$

4 {c m}^{3} / s e c

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(16 / 3) {c m}^{3} / s e c

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(8 / 17) {c m}^{3} / s e c

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None of these

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Solution

The correct option is C $(16 / 3) {c m}^{3} / s e c$
The flow rate of water under a pressure difference is given by
$V = \frac{π r^{4} P}{8 n l} = \frac{8 c m^{3}}{s e c}$
For composite tube
$V = \frac{π r^{4} P}{8 n (3 l / 2)}$
= $\frac{2}{3} \times \frac{π r^{4} P}{8 n l}$
= $\frac{2 \times 8}{3}$
= $\frac{16}{8}$
Since [ $l_{1} = l = 2 l_{2}$ ]

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The correct option is C (16/3)cm3/secThe flow rate of water under a pressure difference is given byV=πr4P8nl=8cm3secFor composite tube V=πr4P8n(3l/2) =23×πr4P8nl =2×83 =168Since [l1=l=2l2]