Two capillary tubes of equal length and inner radii 2r and 4r respectively are added in series and a liquid flows through it. If the pressure difference between the ends of the whole system is 8.5 cm of mercury, find the pressure difference between the two ends of the first capillary tube.

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8. 0 cm of Hg

9.0cm of Hg

10.0 cm of Hg

11.0 cm of Hg

answer is A.

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Detailed Solution

$\frac{P_{1}}{P_{2}} = {(\frac{r_{2}}{r_{1}})}^{4} = {(\frac{4 r}{2 r})}^{4} = \frac{16}{1}$
$\begin{array}{l} P = P_{1} + P_{2} = 17 \to 8.5 cm of Hg; 16 \to ? \\ P_{1} = \frac{16 \times 8 .5}{17} = 8 cm of Hg \end{array}$

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