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Gas laws from KTG

A liquid flow...

Question

A liquid flows through two capillary tubes connected in series. Their lengths are $L$ and $2 L$ and radii $r$ and $2 r$ respectively. Then the pressure differences across the first and the second tube are in the ratio

A

8 : 1

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B

4 : 1

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C

2 : 1

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D

1 : 8

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Solution

The correct option is A $8 : 1$
given :- $l_{2} = 2 l_{1} & r_{2} = 2 r_{1}$
Since flow rate across 1 = flow rate across 2
[as flow rate remains constant]
$\frac{π \times P_{1} r_{1}^{4}}{8 g l_{2}} = \frac{π P_{2} r_{2}^{4}}{8 g l_{2}}$
$\frac{P_{1}}{P_{2}} = \frac{l_{1}}{l_{2}} \times {(\frac{r_{2}}{r_{1}})}^{4} = \frac{1}{2} \times 2^{4} = 8$

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