Fluid pressure

CIE A-Level Physics· 2022-2024 Syllabus, 4(b) Density and pressure· 30 min read

1. Hydrostatic Pressure Law★★☆☆☆⏱ 15 min

In a static fluid, pressure increases with depth because the weight of the fluid above a point pushes down on it. The relationship between pressure, depth and fluid density is derived from balancing the weight of a fluid column against the force from pressure.

📘 Definition

Hydrostatic Gauge Pressure

P_g = \rho g h

Gauge pressure at depth $h$ in a fluid of uniform density $ρ$ is proportional to depth and density. This only accounts for pressure from the fluid, not the atmosphere above the surface.

Example:

At 10 m depth in water, gauge pressure is approximately 1 atm, matching atmospheric pressure at sea level.

📐 Worked Example

Calculate the absolute pressure at a depth of 25 m in seawater (density = 1025 kg m⁻³). Take atmospheric pressure as 1.01 × 10⁵ Pa and $g = 9.81$ m s⁻².

1
First calculate the gauge pressure from the hydrostatic pressure law:
2
$P_g = \rho g h = 1025 \times 9.81 \times 25 = 251206 \text{ Pa}$
3
Add atmospheric pressure to get total (absolute) pressure:
4
$P_{abs} = P_{atm} + P_g = 1.01 \times 10^5 + 2.51 \times 10^5 = 3.52 \times 10^5 \text{ Pa}$

Exam tip:

Always check if the question asks for absolute or gauge pressure before writing your answer.

2. Absolute vs Gauge Pressure★★☆☆☆⏱ 10 min

Most practical pressure gauges (like tire gauges, depth gauges) measure pressure relative to atmospheric pressure, so they output gauge pressure. Exam questions often test your ability to distinguish between the two.

📘 Definition

Gauge Pressure

P_g = P_{abs} - P_{atm}

The difference between total absolute pressure at a point and atmospheric pressure. This is what most mechanical pressure gauges display.

Example:

A tire pressure gauge reads 220 kPa (gauge), so absolute pressure inside the tire is ~320 kPa.

📐 Worked Example

A diver's depth gauge reads 3.2 × 10⁵ Pa (gauge pressure) in fresh water (density = 1000 kg m⁻³). Calculate the diver's depth, taking $P_{a t m} = 1.01 \times 1 0^{5} P a$ .

1
We know the gauge pressure directly from the reading, so use $P_{g} = ρ g h$ :
2
Rearrange to solve for depth $h$ :
3
$h = \frac{P_g}{\rho g} = \frac{3.2 \times 10^5}{1000 \times 9.81} = 32.6 \text{ m}$

✓ Quick check

Test your understanding:

A question gives atmospheric pressure and asks for pressure at the bottom of a lake, with no further specification. What should you calculate?
- Only gauge pressure, because that's what matters
- Absolute pressure, unless the question explicitly asks for gauge
- Either is acceptable in CIE exams
Reveal answer
1 —
CIE exam questions always expect absolute pressure if atmospheric pressure is provided and no specification is given. You will lose marks if you only give gauge pressure.

3. Pascal's Principle and Hydraulic Systems★★★☆☆⏱ 15 min

Pascal's principle describes how pressure is transmitted in enclosed incompressible fluids. This is the working principle behind hydraulic lifts, car brakes and other hydraulic machinery that multiplies force.

📘 Definition

Pascal's Principle

A change in pressure applied to any point of an enclosed incompressible fluid is transmitted undiminished to every point of the fluid and the container walls. This means pressure is equal at the same level in both pistons of a hydraulic system.

Example:

A small force on a small piston creates the same pressure as a large force on a large piston, resulting in force multiplication.

📐 Worked Example

A hydraulic car lift has a small piston with area 0.002 m² and a large piston with area 0.5 m². What force must be applied to the small piston to lift a 1500 kg car on the large piston? Take $g = 9.81$ m s⁻².

1
Calculate the force needed to lift the car (weight of the car on the large piston):
2
$F_2 = mg = 1500 \times 9.81 = 14715 \text{ N}$
3
Apply Pascal's principle (pressure is equal on both pistons):
4
$\frac{F_1}{A_1} = \frac{F_2}{A_2} \implies F_1 = F_2 \times \frac{A_1}{A_2}$
5
Substitute values to find the required input force:
6
$F_1 = 14715 \times \frac{0.002}{0.5} = 58.9 \text{ N}$
7
A force of just ~59 N can lift a 1500 kg car, which demonstrates the force multiplication effect.

4. Common Pitfalls

Wrong move:

Forgetting to add atmospheric pressure when asked for absolute pressure

Why:

Many students default to calculating only gauge pressure from $ρ g h$ and miss the atmospheric term

Correct move:

Always read the question carefully, add $P_{a t m}$ to your $ρ g h$ result if calculating absolute pressure

Wrong move:

Reversing the area ratio in hydraulic system problems

Why:

Confusing which force is applied to which area, or assuming force is equal instead of pressure

Correct move:

Label all variables explicitly before rearranging: write $F_{1} / A_{1} = F_{2} / A_{2}$ with each force/area matched to the correct piston

Wrong move:

Using $g = 10$ m s⁻² when the question expects $9.81$ m s⁻²

Why:

Early rounding or approximation changes the final result beyond the allowed tolerance in mark schemes

Correct move:

Always use $g = 9.81$ m s⁻² unless the question explicitly tells you to use another value

Wrong move:

Assuming all water has density 1000 kg m⁻³

Why:

Seawater has a higher density than fresh water, and questions will specify the density to use

Correct move:

Always use the density value given in the question, don't just default to 1000 kg m⁻³

5. Quick Reference Cheatsheet

Concept	Formula	Key Notes
Gauge pressure at depth $h$	$P_{g} = ρ g h$	Pressure from fluid only, no atmosphere
Absolute pressure	$P_{ab s} = P_{a t m} + P_{g}$	Total pressure at depth
Gauge pressure definition	$P_{g} = P_{ab s} - P_{a t m}$	Output of most pressure gauges
Pascal's principle (hydraulics)	$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$	Pressure is equal, force scales with area

When this came up on past exams

AI-estimated based on syllabus patterns — cross-check with official past papers for accuracy. Use only as revision-focus signals.

2023 · 13
Calculate depth from given pressure
2022 · 12
Hydraulic lift force calculation
2021 · 22
Absolute vs gauge pressure problem

Going deeper

syllabusCIE 9702 Official SyllabusCheck for latest updates

What's Next

Fluid pressure is a core foundational concept for all further fluid mechanics topics in CIE A-Level Physics. It is directly required to understand upthrust and Archimedes' principle, which is a very common exam topic that often combines pressure, force and density concepts. Fluid pressure also underpins ideas about fluid flow and buoyancy, which can appear in combined practical and structured questions. Mastering the calculations and distinctions covered here will help you secure easy marks on multiple choice and short structured questions, which are common in both Paper 1 and Paper 2.