2022 A/L Physics Past Paper · Question 14 (English Medium)

Answer: (5) $1.6 \times 10^{7}\ \text{Pa}$

Given information

What we know:

The nipple is a narrow tube. Its length is $l = 3 \times 10^{-3}\ \text{m}$.

Its diameter is $d = 5 \times 10^{-4}\ \text{m}$.

The coefficient of viscosity of the grease is $\eta = 80\ \text{Pa s}$.

The volume pushed through is $V = 10^{-6}\ \text{m}^3$ in a time $t = 10\ \text{s}$.

We may take $(2.5)^4 = 40$ and $\pi = 3$.

What the question asks

We want: the pressure difference $\Delta p$ between the two ends of the nipple.

What each letter means:

$V$ ($\text{m}^3$): the volume of grease flowing through.

$t$ ($\text{s}$): the time taken.

$a$ ($\text{m}$): the radius of the tube.

$l$ ($\text{m}$): the length of the tube.

$\eta$ ($\text{Pa s}$): the coefficient of viscosity of the liquid.

$\Delta p$ ($\text{Pa}$): the pressure difference between the two ends.

Relevant theory

Poiseuille's equation provides an expression for the rate of flow of a liquid in uniform

flow through a capillary tube.

The coefficient of the viscosity of a fluid is defined as the tangential force on a unit area

under a unit velocity gradient of the fluid in uniform stream line motion.

The tube should be horizontal and narrow.

For example, think of pushing honey through a thin straw.

Thick honey needs a harder push than thin juice. A longer straw needs a harder push. A thinner straw needs a much harder push.

The book puts this into one equation. The push $\Delta p$ must be larger for a bigger $\eta$ and a longer tube $l$. It must be smaller when the radius $a$ is bigger, and the radius counts to the fourth power.

$$\dfrac{V}{t} = \dfrac{\pi a^4}{8\eta}\left(\dfrac{\Delta p}{l}\right)$$

In words: the volume flowing each second equals $\pi a^4$ times the pressure difference per unit length, divided by $8\eta$.

Required equations

$\dfrac{V}{t} = \dfrac{\pi a^4}{8\eta}\left(\dfrac{\Delta p}{l}\right)$

$\Delta p = \dfrac{8\eta l}{\pi a^4}\cdot\dfrac{V}{t}$

Applying it to this question

All the quantities are in SI units: metres, seconds, pascal seconds. So $\Delta p$ comes out in pascals.

$2.5e-4$

$a = \dfrac{5 \times 10^{-4}}{2}$

$4e-15$

$a^4 = (2.5)^4 \times 10^{-16} = 40 \times 10^{-16}$

$1e-7$

$\dfrac{V}{t} = \dfrac{10^{-6}}{10}$

$1.92$

$8 \eta l = 8 \times 80 \times 3 \times 10^{-3}$

$1.6e7$

$\Delta p = \dfrac{1.92}{3 \times 4 \times 10^{-15}} \times 10^{-7} = \dfrac{1.92 \times 10^{-7}}{1.2 \times 10^{-14}}$

Reasoning and calculation

The tube is narrow, so the radius to the fourth power is tiny, $4 \times 10^{-15}\ \text{m}^4$. That makes the pressure difference large.

The pressure difference is $1.6 \times 10^{7}\ \text{Pa}$. That is option (5).

Checks

$8 * eta * l * (V / t) / (3 * q)$

Dimension check: $\text{Pa s} \times \text{m} \times \text{m}^3\ \text{s}^{-1} \div \text{m}^4 = \text{Pa}$. That is a pressure, correct.

Does it make sense?

Put the answer back: the volume pushed through in $10\ \text{s}$ is $6.25 \times 10^{-14} \times 1.6 \times 10^{7} = 10^{-6}\ \text{m}^3$. That is what the question gave.

$1.6 \times 10^{7}\ \text{Pa}$ is about $160$ times the air pressure around us. A hand grease gun does give pressures like that, because the nipple is so narrow.

Conclusion

$\Delta p = 1.6 \times 10^{7}\ \text{Pa}$

The pressure difference is $\Delta p = \dfrac{8\eta l}{\pi a^4}\cdot\dfrac{V}{t} = 1.6 \times 10^{7}\ \text{Pa}$.

The answer is (5).

Use the radius, not the diameter, in Poiseuille's equation. Halve the diameter first.

Put your answer back into $\dfrac{V}{t}$ to check the power of ten.

Exam tip