Water flows in a 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 a/2 and the pressure increased to 2P, the rate of flow becomes

The correct answer is Q/8.

Step-by-Step Explanation:

According to Poiseuille's law, the rate of flow $Q$ of a viscous fluid flowing in a streamlined manner through a capillary tube is given by the formula:

$Q=\frac{\pi P{r}^4}{8\eta l}$

where:
- $P$ is the pressure difference across the capillary tube,
- $r$ is the radius of the capillary tube,
- $\eta$ is the coefficient of viscosity of the fluid, and
- $l$ is the length of the capillary tube.

From Poiseuille's formula, for a given liquid and length of tube, the volumetric flow rate is directly proportional to the pressure difference $P$ and the fourth power of the radius $r$:

$Q\propto P{r}^4$

Initial state:
Radius $r_1=a$
Pressure difference $P_1=P$
Rate of flow $Q_1=Q$

Final state:
Radius $r_2=\frac{a}{2}$
Pressure difference $P_2=2P$
Let the new rate of flow be $Q_2$.

Taking the ratio of the new flow rate to the initial flow rate:

$\frac{Q_2}{Q_1}=\left(\frac{P_2}{P_1}\right)\times {\left(\frac{r_2}{r_1}\right)}^4$

Substitute the given values into the ratio:

$\frac{Q_2}{Q}=\left(\frac{2P}{P}\right)\times {\left(\frac{a/2}{a}\right)}^4$

$\frac{Q_2}{Q}=2\times {\left(\frac{1}{2}\right)}^4$

$\frac{Q_2}{Q}=2\times \frac{1}{16}=\frac{1}{8}$

$Q_2=\frac{Q}{8}$

Hence, the new rate of flow becomes Q/8.