Question
A $16cm^3$ volume of water flows per second through a capillary tube of radius r сm and of length 1cm when connected to a pressure head of h cm of water. If a tube of the same length and radius $\frac{\text{r}}{2}$ is connected to the same pressure head, find the mass of water flowing per minute through the tube.

Answer

Here, $V_1 = 16cm^3/ \sec$;$\text{P}_1=\text{h}\rho\text{g};\text{r}_1=\text{r};\text{l}_1=\text{l} $
$\text{V}_2=?;\text{l}_2=\text{l};\text{r}_2=\frac{\text{r}}{2};\text{P}_2=\text{h}\rho\text{g};$
So, $\text{P}_1=\text{P}_2$ Now, $\text{V}_1=\frac{\pi\text{P}_1\text{r}^4_1}{8\eta\text{l}_1}$ and $\text{V}_2=\frac{\pi\text{P}_2\text{r}^4_2}{8\eta\text{l}_2}$$\therefore\frac{\text{V}_2}{\text{V}_1}=\frac{\text{P}_2}{\text{P}_2}\times\frac{\text{r}^4_2}{\text{r}^4_1}\times\frac{\text{l}_1}{\text{l}_2}$
$=\Big(\frac12\Big)=\frac{1}{16}$
$\text{V}_2=\frac{16}{16}=1\text{cm}^3/\text{s}$
Volume of water flowing per minute,$=1\times60$
$=60\text{cm}^3/\text{min}$
$\therefore$ Mass of water flowing per minute,
= 60gm

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