Two capillary tubes of lengths $15 c m$ and $5 c m$ and radii $0.06 c m$ and $0.02 c m$ respectively are connected in series. If the pressure difference across the end faces is equal to pressure of $15 c m$ high water column, then the pressure difference across the (a) first tube and (b) second tube are

0.536 c m

of water column;

14.464 c m

of water column

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0.536 m

of water column;

14.464 m

of water column

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5.36 c m

of water column;

14.464 c m

of water column

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5.36 m

of water column;

14.464 m

of water column

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Solution

The correct option is A $0.536 c m$ of water column; $14.464 c m$ of water column
Let the Pressure at the function be $P_{1}$ cm high water column.
as, the pipes are in series, the rate water flowing through the pipes will be same,
$\to \frac{π (P - P_{1}) (t g) (0.06)^{4}}{8 η (15) 3} = \frac{π (P_{1}) (0.02)^{4} (t g)}{8 η (5)}$
$\to 27 P = 18 P_{1} \Rightarrow P_{1} = \frac{27 P}{28}$
So, pressure difference across first pipe
$= \frac{15}{28} c m = 0.536$ cm of water column
b)Second tube= $\frac{27}{28} \times 15 c m = 14.464 c m$ of water column.

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