Question

Water flows in streamlined manner through a capillary tube of radius a, the pressure difference being P and the rate of flow Q. If the radius is reduced to $\frac{a}{2}$ and the pressure increased to 2P, the rate of flow becomes

• A. $4Q$
• B. $Q$
• C. $\frac{Q}{4}$
• D. $\frac{Q}{8}$

Answer 1

The correct option is D $\frac{Q}{8}$

$V=\frac{\pi P{r}^4}{8\eta l}\therefore V\propto P{r}^4$ [$\eta$ and l are constants]
$\therefore \frac{V_2}{V_1}=\left(\frac{P_2}{P_1}\right){\left(\frac{r_2}{r_1}\right)}^4=2\times {\left(\frac{1}{2}\right)}^4=\frac{1}{8}\therefore V_2=\frac{Q}{8}$.