Physics 0625 · syllabus 1.8
Cambridge IGCSE Physics 0625 — Pressure topical past papers
Pressure as force per unit area, everyday examples of high and low pressure, and how pressure beneath a liquid surface grows with depth and density, with Δp = ρgΔh.
- Questions
- 756
- Years
- 2009–2026
- With mark scheme
- 99%
- Questions 2022–2026
- 199
Anyone can browse these pages. Opening the questions and their mark schemes takes a free Momentum account.
What Pressure covers
Pressure is the force acting on each unit of area, p = F/A, measured in pascals (1 Pa = 1 N/m²) or in N/cm². The same force gives a large pressure on a small area and a small pressure on a large one. That is why a sharp knife cuts, a drawing pin goes into wood with a gentle push, and a camel's wide feet, snowshoes and the broad tyres of a tractor stop them sinking into soft ground.
In a liquid the pressure increases with depth, because there is more liquid above pushing down, and at the same depth it is greater in a denser liquid. That is why a dam is built thicker at its base, and why water spurts from a hole near the bottom of a tank faster than from one near the top. At any one depth the pressure acts in all directions, and it is the same whatever the shape of the container.
Extended candidates calculate the change in pressure beneath a liquid surface with Δp = ρgΔh, where ρ is the density of the liquid and Δh the change in depth. Pressure questions often begin with a weight, which supplies the force, or with a density, and gas pressure returns in thermal physics, where it is explained by particles colliding with the walls of their container.
In the syllabus
The headings this topic has in the Cambridge International 2026–2028 syllabus, as the board prints them.
- 1.8 Pressure
What you should be able to do
- Use p = F/A, converting cm² to m² where needed
- Explain everyday examples in which a small area raises the pressure or a large area lowers it
- Describe how the pressure in a liquid depends on the depth and on the liquid's density
- Use Δp = ρgΔh to find a pressure change, a depth or a density (Extended)
- Find the force a pressure exerts on a surface, such as water pressing on a window or a dam
Key ideas
- Pressure p = F/A; 1 Pa = 1 N/m²
- 1 m² = 10 000 cm², so 1 N/cm² = 10 000 Pa
- Pressure in a liquid increases with depth and with the density of the liquid, whatever the shape of the container
- Change in pressure beneath a liquid surface Δp = ρgΔh (Extended)
How Pressure has been examined
From 2022 to 2026, Pressure was on Paper 3 · Theory (Core) in 13 of 14 exam series, worth about 9% of that paper’s marks.
| Paper | Questions | Share of marks | Series |
|---|---|---|---|
| Paper 1 · Multiple Choice (Core) | 83 | 6% | 14 of 14 |
| Paper 2 · Multiple Choice (Extended) | 64 | 5% | 14 of 14 |
| Paper 3 · Theory (Core) | 28 | 9% | 13 of 14 |
| Paper 4 · Theory (Extended) | 21 | 6% | 12 of 14 |
| Paper 6 · Alternative to Practical | 3 | 3% | 2 of 14 |
Momentum’s own count, over the questions in its bank for these years: a series is one exam session, every zone included, and a question filed under two chapters counts in both. Median marks per question: 1. The bank files this topic together with neighbouring ones, so these figures count every question in its chapters, including some on those topics.
Filed in Momentum under: Forces and Pressure.
A worked example
A diver is 12 m below the surface of seawater of density 1030 kg/m³. Find the pressure due to the water at that depth, and the force it exerts on the glass of her mask, which has an area of 0.015 m². (g = 9.8 N/kg)
- Pressure due to the water: Δp = ρgΔh = 1030 × 9.8 × 12 = 121 128 Pa, about 1.2 × 10⁵ Pa
- Force on the glass: F = pA = 121 128 × 0.015 = 1817 N
- Check the unit: Pa × m² = (N/m²) × m² = N
Answer: 1.2 × 10⁵ Pa; a force of about 1.8 × 10³ N
Written by Momentum for this page, with its own numbers; it is not taken from a past paper.
Mistakes that cost marks
- Converting cm² to m² by dividing by 100 — there are 10 000 cm² in 1 m², so 25 cm² is 0.0025 m²
- Using a mass as the force — a 60 kg person standing on 0.040 m² exerts 60 × 9.8 ÷ 0.040 = 14 700 Pa, not 1500 Pa
- Thinking a wider container gives a greater pressure at the bottom — at the same depth in the same liquid the pressure is the same, whatever the container's shape
- Using the density of a submerged object in Δp = ρgΔh — the pressure comes from the liquid above the point, so ρ is the liquid's density
- Measuring the depth up from the bottom of the container — the depth is measured down from the liquid surface
Exam technique
- Write p = F/A, rearrange it first, and show any conversion of the area on a line of its own
- In everyday examples, link area and pressure in one sentence: 'the area is small, so for the same force the pressure is large'
- If a question gives atmospheric pressure and asks for the total pressure at a depth, add it to ρgΔh; if it asks for the pressure due to the liquid, leave it out
- Check that depths are in metres and densities in kg/m³, so that ρgΔh comes out in pascals
Practising in Momentum
- Every question in a chapter as one list, from every paper and year in the bank, newest first or shuffled.
- Each question turns over to its own part of the mark scheme, where the board published one.
- Narrow by paper, year, session, zone and marks, and star the questions worth keeping.
- Pick chapters, a number of questions and a time, and the exam builder draws a mock to sit and mark.
Past-paper questions and mark schemes are the copyright of Cambridge International. Momentum shows them to signed-in learners for revision and does not publish them on these pages.