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Two capillary tubes of length 15cm and 5cm and radii 0.06 cm and 0.02 cm respectively are connected in series. If the pressure difference across the end faces is equal to pressure of 14cm high water column then find the pressure difference across the (i) first tube and ( ii ) second tube

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3.5, 10.5 cm of water column

0.5,13.5 cm of water column

2.54, 11.5cm of water column

1.5,12.5cm of water column

answer is A.

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Detailed Solution

$R = R_{1} + R_{2}; R = \frac{8 η}{π} (\frac{l_{1}}{r_{1}^{4}} + \frac{l_{2}}{r_{2}^{4}}); V = \frac{P}{R_{1} + R_{2}}$

$Here P = P_{1} - P_{2} = (P_{1} - P_{0}) + (P_{0} - P_{2})$
$P_{1} - P_{0} and P_{0} - P_{2}$ are pressure difference across individual capillaries $\frac{P_{1} - P_{0}}{P_{0} - P_{2}} = \frac{ℓ_{1}}{ℓ_{2}} {(\frac{r_{2}}{r_{1}})}^{4}$

$\frac{P_{1} - P_{0}}{P_{0} - P_{2}} = \frac{1}{27}$

$P_{1} - P_{0} = \frac{14}{28} \times 1 = 0.5$

$P_{0} - P_{2} = \frac{14}{28} \times 27 = 13.5$

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