2019 A/L Physics Past Paper · Question 28 (English Medium)

Answer: (5) $\dfrac{125}{4\pi}$

Given information

We know:

The heart pumps $Q = 5\ \text{L}$ of blood every minute.

There are $N = 10^9$ capillaries.

Each one is $8\ \mu\text{m}$ wide. So its radius is $4\ \mu\text{m}$.

What the question asks

We want:

The average speed of the blood in the capillaries, in $\text{cm min}^{-1}$.

Relevant theory

Volume of fluid entering the tube through area $A_1$.

Since the flow is continuous, the volume of fluid flowing past the cross section $A_2$ per unit is the same.

This equation which explains the variation of velocity of flow with the variation of cross section of the tube of flow, is referred to as the equation of continuous flow or the equation of continuity.

Required equations

$Q = A v$

$A_{tot} = N\pi r^2$

$v = \frac{Q}{N\pi r^2}$

Applying it to this question

The plan:

Write everything in cm and minutes.

Find the area of one capillary.

Find the total area of all $10^9$.

The same blood each minute goes through this total area.

So $v = \dfrac{Q}{A_{tot}}$.

Reasoning and calculation

$5000$

$Q = 5\ \text{L min}^{-1} = 5000\ \text{cm}^3\text{ min}^{-1}$

$4 \times 10^{-4}$

$r = 4\ \mu\text{m} = 4 \times 10^{-4}\ \text{cm}$

$16\pi \times 10^{-8}$

$a = \pi (4 \times 10^{-4})^2 = 16\pi \times 10^{-8}$

$160\pi$

$A_{tot} = 10^9 \times 16\pi \times 10^{-8}$

$v = \frac{5000}{160\pi}$

$9.95$

$v = \frac{125}{4\pi}$

Checks

Check the unit.

$Q$ is in $\text{cm}^3\text{ min}^{-1}$.

$A$ is in $\text{cm}^2$.

So $v$ is in $\text{cm min}^{-1}$.

Does it make sense?

The blood moves only about $10\ \text{cm}$ every minute.

That is very slow.

The reason: all the capillaries together have a very wide area, about $500\ \text{cm}^2$.

The slow blood has time to give food and air to the body.

$Q / (N * pi * r^2)$

Conclusion

$v = \frac{125}{4\pi}\ \text{cm min}^{-1}$

$$v = \frac{5000}{160\pi} = \frac{125}{4\pi}\ \text{cm min}^{-1}$$

So the answer is option (5).