Question

A horizontal composite capillary tube has a radius $2r$ for a length $2L$ and radius $r$ for a length $L$ as shown and is connected to a tank at one end and left free at the other end. The tank contains a liquid of coefficient of viscosity $\eta$ . If a constant pressure difference $P$ exists across the ends of the capillary tube, the volume flux through the capillary tube is

• A. $\left(\frac{16}{17}\frac{\pi P{r}^4}{8\eta L}\right)$
• B. $\left(\frac{9}{8}\frac{\pi P{r}^4}{8\eta L}\right)$
• C. $\left(\frac{17}{16}\frac{\pi P{r}^4}{8\eta L}\right)$
• D. $\left(\frac{8}{9}\frac{\pi P{r}^4}{8\eta L}\right)$

Answer 1

The correct option is D $\left(\frac{8}{9}\frac{\pi P{r}^4}{8\eta L}\right)$

Let the $P$ in Middle be $P_1$,

So, As the pipes are in series the flow rate should be same in both pipe for conyinuity,

$\frac{\pi (P-P_1{r}_2{)}^4}{8\eta 2}=\frac{P_1{r}^4}{8\eta }$

$\to \frac{16(P-P_1)}{2}=P_1\Rightarrow 8P-8P_1=P_1\Rightarrow \frac{8P}{9}=P_1$

So, The flux (Volume flow rate ) through the capillary is

$\frac{\pi P_1{r}^4}{8\eta L}=\frac{8}{9}\left(\frac{\pi P{r}^4}{8\eta L}\right)$