Pressure

This lesson explains Pressure for Cambridge IGCSE Physics 0625. It separates the Core requirements from the additional Supplement work for Extended candidates. Pay close attention to equation choice, unit conversion and the final sense check. Read the explanation once, then attempt the worked method from a blank page.

What is the pressure equation?

Pressure is the force per unit area. In words: pressure = force ÷ area. In symbols: $p = \dfrac{F}{A}$. The unit is the pascal (Pa), where $1\ \text{Pa} = 1\ \text{N/m}^2$. The same force on a smaller area gives a larger pressure: a drawing pin concentrates your push onto a tiny point, while skis spread your weight to stop you sinking into snow.

Quantity	Symbol	Unit
Pressure	$p$	Pa (N/m²)
Force	$F$	N
Area	$A$	m²
Density	$\rho$	kg/m³
Gravitational field strength	$g$	9.8 N/kg
Depth change	$\Delta h$	m

How does pressure change with depth in a liquid? (Extended)

The change in pressure equals density × gravitational field strength × change in depth. In symbols: $\Delta p = \rho g \Delta h$. This part is Extended (Supplement) only. Three facts earn marks: pressure in a liquid increases with depth, increases with density, and acts equally in all directions at a given depth. That is why dam walls are thicker at the bottom and why a diver’s ears hurt more the deeper they go. Total pressure under water $=$ atmospheric pressure $+\ \rho g \Delta h$; read carefully whether the question wants the total or just the liquid’s contribution.

Worked example

A reservoir behind a dam holds fresh water of density 1000 kg/m³. Use $g = 9.8\ \text{N/kg}$.

(a) Calculate the pressure due to the water at a depth of 12 m. [3] (b) An inspection hatch at this depth has an area of 0.50 m². Calculate the force on the hatch due to the water. [2]

Solution (a). Equation: $\Delta p = \rho g \Delta h$. Substitute: $\Delta p = 1000 \times 9.8 \times 12$. Answer: $\Delta p = \textbf{117 600 Pa} \approx \mathbf{1.2 \times 10^{5}}\ \textbf{Pa}$.

Solution (b). Equation: $p = \dfrac{F}{A}$, rearranged to $F = pA$. Substitute: $F = 117\,600 \times 0.50.$ Answer: $F = \textbf{58 800 N} \approx \mathbf{5.9 \times 10^{4}}\ \textbf{N}$.

Original marking points

• M1: $\Delta p = \rho g \Delta h$ stated or used.
• M1: correct substitution $1000 \times 9.8 \times 12$.
• A1: $117\,600\ \text{Pa}$ (accept $1.2 \times 10^{5}\ \text{Pa}$) with unit.
• M1: $F = pA$ with their pressure $\times\ 0.50$ (error carried forward allowed).
• A1: $58\,800\ \text{N}$ (accept $5.9 \times 10^{4}\ \text{N}$).

These marking points belong to this original example. They are not an official Cambridge mark scheme.

Common errors and how to correct them

• Leaving area in cm². $1\ \text{m}^2 = 10\,000\ \text{cm}^2$, so the error is huge. Fix: convert to m² before substituting; divide cm² by 10 000.
• Inverting the equation. $A/F$ gives pressure falling as force rises, which is nonsense. Fix: bigger force, bigger pressure; check your answer’s direction of change.
• Forgetting atmospheric pressure. When asked for total pressure at depth, add about $1.0 \times 10^{5}\ \text{Pa}$. Fix: underline “total” or “due to the liquid” in the question.
• Using weight in kg. $F$ must be in newtons. Fix: weight $= mg$ with $g = 9.8\ \text{N/kg}$ (some papers use 10, so read the question).
• Thinking depth pressure depends on container shape or width. It depends only on $\rho$, $g$ and $\Delta h$. Fix: a narrow tube and a lake give equal pressure at equal depth.

How to apply this in an exam

Before any $p = \dfrac{F}{A}$ substitution, write the area conversion as its own line: "$A = 20\ \text{cm}^2 = 20 \div 10\,000 = 0.0020\ \text{m}^2$". Mark schemes frequently allocate a mark to the converted value itself, and showing it protects the later answer mark too. Writing the conversion separately makes the method visible and reduces a recurring area-conversion error.

Where this skill matters

Both tiers calculate $p = \dfrac{F}{A}$ on Papers 1/3 (Core) and 2/4 (Extended); MCQs favour blocks resting on different faces, where the same weight gives different pressures. Only Extended papers test $\Delta p = \rho g \Delta h$, usually as a 2-3 mark calculation like the dam question, sometimes linked to a mercury barometer or manometer. Paper 5/6 relevance is indirect: pressure ideas appear in liquid-column and density practicals. Pressure also returns in Thermal Physics as particle collisions on container walls, so revise the two together for efficient coverage.