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Water flows through a capillary tube of radius $' r'$ and length at a rate of $40 m l$ per second, when connected to a pressure difference of $' h' c m$ of water. Another tube of the some length but radiud . $r / 2$ is connected in series with this tube and the combination is connected to the same pressure head. [density of water is $ρ$ ]

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The pressure difference across each tube is $P_{1} = \frac{ρ g h}{17}$ and $P_{2} = \frac{16}{17} ρ g h$

The pressure difference across each tube is $P_{1} = \frac{ρ g h}{16}$ and $P_{2} = \frac{17}{16} ρ g h$

The rate of flow of water through the combination is $\frac{40}{17} c . c / s e c$ .

The rate of flow of water through the combination is $\frac{17}{40} c . c / s e c$

answer is A, C.

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Detailed Solution

According to Poiseuille’s equation.

Rate of flow= $\frac{P}{R}$ where P is the pressure difference between the ends of pipe .

Making an analogy with current flow,

Let $R_{1} = \frac{8 η L}{π r^{4}}$ ; $R_{2} = \frac{8 η L}{π {(\frac{r}{2})}^{4}} = 16 R_{1}$

$P_{1} + P_{2} = P = ρ g h$ and $P_{1} / P_{2} = \frac{R_{1}}{R_{2}} = \frac{1}{16}$

Also Rate of flow= $\frac{P}{R_{1} + R_{2}} = \frac{P}{17 R_{1}} = \frac{40}{17} = m L / s$

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