IGCSE · 26 September 2026 · 6 min read
Density and pressure questions: three IGCSE worked examples
Density and pressure questions share enough symbols to tempt students into using whichever equation they remember first. Their meanings are different: density describes mass per volume, while pressure describes perpendicular force per area. A liquid-pressure calculation connects them through the weight of the liquid above a point.
These original examples focus on the modelling decision and the unit conversions that control the result. They support relevant Cambridge and Edexcel IGCSE topics; check your course's exact scope for quantitative liquid-pressure work.
Name the physical quantity before selecting a formula
Ask what the question compares. If it gives a sample's mass and occupied volume, density is the likely connection. If it gives a perpendicular force spread across a contact area, use pressure. If it asks how pressure changes below a liquid surface, identify density, gravitational field strength and vertical depth.
Write the chosen relationship in words once if the symbols feel confusing. In p = ρgh, p is pressure and the Greek letter ρ represents density. They are not the same quantity. Include the output unit in your plan: kg/m³ for SI density, pascals for pressure. That makes an accidental substitution easier to spot.
Worked example: density from displaced volume
A metal object has mass 156 g. Water in a measuring cylinder rises from 40 cm³ to 60 cm³ when the object is fully submerged. The object's volume is the change, 20 cm³, not the final cylinder reading. Its density is 156 / 20 = 7.8 g/cm³, equivalent to 7800 kg/m³.
For an SI calculation, convert mass to 0.156 kg and volume to 20 × 10⁻⁶ m³ = 2.0 × 10⁻⁵ m³. Then ρ = 0.156 / (2.0 × 10⁻⁵) = 7800 kg/m³. Check for trapped bubbles or incomplete submersion in a real measurement: either can make the displacement an inaccurate representation of the object's volume.
Worked example: contact pressure and a changed orientation
A 12 kg block rests on a horizontal surface over a contact area of 200 cm². Use g = 10 N/kg as supplied for this example. The downward force is its weight, 120 N. The area is 200 × 10⁻⁴ = 0.020 m², so pressure = 120 / 0.020 = 6000 Pa.
Turn the block so the contact area becomes 100 cm² while its weight remains unchanged. The pressure doubles to 12 000 Pa. This is a useful prediction before calculation: spreading the same force over a smaller area increases force per unit area. Dividing 12 by 200 would mix mass with force and cm² with an SI pressure label.
Worked example: pressure below a liquid surface
Find the pressure increase 0.80 m below the surface of oil with density 850 kg/m³, using g = 10 N/kg. The increase is Δp = ρgh = 850 × 10 × 0.80 = 6800 Pa. If the surface is open to air at a stated pressure of 101 000 Pa, the total pressure there is 107 800 Pa.
Now double the depth to 1.60 m in the same oil. The liquid's pressure contribution doubles to 13 600 Pa, but the total pressure including the unchanged atmospheric contribution does not double. Keep these two quantities separate. A question asking for pressure due to the liquid does not require adding atmospheric pressure.
Diagnose the square and cube conversion traps
A centimetre is one hundredth of a metre. A square centimetre is therefore (0.01)² m² = 10⁻⁴ m², while a cubic centimetre is (0.01)³ m³ = 10⁻⁶ m³. Using the length conversion directly for an area or volume changes the physical size by the wrong factor.
You can sometimes calculate in consistent non-SI units first, as the density example did, but label them honestly and convert only when required. A result of 7.8 g/cm³ is not 7.8 kg/m³. Check an answer by asking whether the material, force or depth trend is plausible, rather than trusting a neat-looking calculator display.
Independent practice: choose the model yourself
A stone of mass 90 g displaces 30 cm³ of water. Its density is 3.0 g/cm³, or 3000 kg/m³. A separate 50 N force acts over 25 cm²: the area is 0.0025 m², giving pressure 20 000 Pa. These calculations use different equations even though both involve dividing one quantity by another.
Finally, compare two points at the same vertical depth in the same stationary liquid under the same surface pressure. They have the same pressure even if one container is wider. The total force on a base can still differ because force equals pressure multiplied by area. Write that distinction in your correction log if you tend to treat pressure and total force as synonyms.
Questions, explained
Choose a question for a direct answer, then explore the explanation and supporting resources. Each answer has its own link to save or share.
What is the difference between density and pressure?
Density is mass per unit volume and has SI unit kg/m³. Pressure is perpendicular force per unit area and has SI unit Pa, equivalent to N/m². A liquid's density affects how pressure increases with depth, but the two quantities are not interchangeable.
How do I calculate density using a measuring cylinder?
For a suitable object fully submerged in the liquid, subtract the initial volume reading from the final reading to obtain displaced volume. Divide the object's mass by that volume using consistent units. Avoid trapped air, splashing and using the final cylinder reading as the object's volume.
How do I convert cm² to m² for pressure?
Multiply the area in cm² by 10⁻⁴. For example, 200 cm² = 0.020 m². The length conversion must be squared because area has two dimensions. Using 10⁻² for an area is a different conversion and gives an incorrect pressure.
When should I add atmospheric pressure to ρgh?
Add the stated surface pressure when the question asks for total pressure at depth. The quantity ρgh gives the increase due to the liquid above that point. If the question asks only for pressure due to the liquid, that increase is the required quantity.
Does a wider container mean greater liquid pressure?
At the same vertical depth in the same stationary liquid with the same surface pressure, pressure is the same. A wider base can experience a larger total force because F = pA. Container width and total liquid volume do not replace depth in the liquid-pressure relationship.