What is pressure in the context of Physics, and how is it measured?

In Physics, pressure is defined as the force exerted per unit area on the surface of an object. It is a scalar quantity and is measured in pascals (Pa) in the International System of Units (SI). One pascal is equivalent to one newton per square meter. Understanding pressure is crucial in the Cambridge IGCSE curriculum as it applies to various real-world scenarios, such as atmospheric pressure and pressure in fluids.

How does pressure relate to force and area in Physics?

Pressure is directly related to force and inversely related to the area over which the force is applied. The formula for pressure is P = F/A, where P is pressure, F is the force applied, and A is the area. This relationship means that for a constant force, increasing the area over which the force is distributed will decrease the pressure, and vice versa. This concept is fundamental in the study of forces and energy in the Cambridge IGCSE Physics syllabus.

What are some common examples of pressure in everyday life?

Pressure is a concept that we encounter frequently in everyday life. Some common examples include the pressure exerted by the atmosphere on the Earth's surface, the pressure in car tires, and the pressure applied by a sharp knife which makes it easier to cut through objects. Understanding these examples helps students relate the theoretical aspects of pressure to practical situations, which is an important aspect of the Cambridge IGCSE Physics curriculum.

How does atmospheric pressure change with altitude, and why is this important?

Atmospheric pressure decreases with an increase in altitude. This is because the density of air molecules decreases as you move higher above the Earth's surface, resulting in less force exerted over a given area. This concept is important in the study of Physics as it affects various phenomena such as weather patterns and the functioning of aircraft. It is a key topic in the Cambridge IGCSE Physics curriculum under the study of motion, forces, and energy.

What is the difference between absolute pressure and gauge pressure?

Absolute pressure is the total pressure exerted on a system, including atmospheric pressure, whereas gauge pressure is the pressure relative to atmospheric pressure. Gauge pressure is often used in practical applications, such as measuring tire pressure, where the atmospheric pressure is not included in the measurement. Understanding the distinction between these two types of pressure is essential for students studying the Cambridge IGCSE Physics curriculum, as it helps in solving problems related to pressure in various contexts.

Why is "Using $\text{cm}^{2}$ in $p = F/A$" a common mistake in Pressure?

Why it happens: Reading the diagram in cm. How to avoid it: Convert area to m^2 for the answer in pascals: 1\,m^2 = 10,000\,cm^2. Source: 0625/42 — recurring

Why is "Using horizontal distance in $\rho g h$" a common mistake in Pressure?

Why it happens: Misreading the diagram. How to avoid it: h is VERTICAL depth below the free surface.

Why is "Mixing g/cm³ with kg/m³ in $\rho g h$" a common mistake in Pressure?

Why it happens: Different unit systems in textbook tables. How to avoid it: Use SI: in kg/m³, h in m, g in N/kg, then p comes out in Pa.

Why is "Forgetting atmospheric pressure when 'total' is asked" a common mistake in Pressure?

Why it happens: Computing only g h. How to avoid it: Total = atmospheric + fluid contribution. Read the question carefully.

Motion, Forces and EnergyCambridge IGCSE Physics (0625)

Pressure

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Pressure — Cambridge IGCSE 0625 Physics Extended (2026)

Force per unit area, plus the special $p = h\rho g$ formula for pressure in a fluid. The hydraulic-press question is a Cambridge favourite.

What you’ll learn

Mapped to the Cambridge IGCSE 0625 syllabus (2026-2028).

1.8 — Recall and use $p = F/A$ .
1.8 — Recall and use $p = h\rho g$ for the change in pressure with depth in a fluid.
1.8 — Describe how pressure is transmitted through a fluid in hydraulic systems.

Pressure as force per area

$p = F/A$ . Same force, smaller area, more pressure.

Formula. $p = \dfrac{F}{A}.$

$p$ : pressure ( $\text{Pa}$ ).
$F$ : force ( $\text{N}$ ).
$A$ : area ( $\text{m}^{2}$ ).

$1\,\text{Pa} = 1\,\text{N/m}^{2}$ .

Worked. A $600\,\text{N}$ person stands on snow shoes (area $0.4\,\text{m}^{2}$ total).

$p = 600/0.4 = 1500\,\text{Pa}$ .
Without snow shoes (area $0.04\,\text{m}^{2}$ ): $p = 600/0.04 = 15{,}000\,\text{Pa}$ — sinks into snow.

Sharp = high pressure. A drawing pin: small contact area on the wall side, large area on the thumb side. Same force on both ends, but the wall side has tiny area → huge pressure → pin pierces.

Reducing pressure. Snowshoes, tank tracks, snow tyres — all spread weight over more area.

The same force on a smaller area produces a much higher pressure.

$p = F/A$ .
$\text{Pa} = \text{N/m}^{2}$ .
Smaller area → higher pressure for same force.
Sharp objects exploit small area.

Pressure in a fluid

$p = h\rho g$ . Pressure rises with depth and density.

Formula. $p = h\rho g,$ where $h$ is the depth below the surface, $\rho$ is the fluid density, $g$ is gravitational field strength.

This gives the pressure DUE TO THE FLUID at that depth. To get total pressure including atmosphere: add atmospheric pressure ( $\sim 100{,}000\,\text{Pa}$ ).

Worked. Pressure $5\,\text{m}$ below the surface of seawater ( $\rho = 1030\,\text{kg/m}^{3}$ ).

$p = 5 \times 1030 \times 9.8 \approx 50{,}500\,\text{Pa}$ (just from the water).

Why pressure increases with depth. A column of fluid above presses down. Deeper = more fluid above = more weight per area below.

All directions. At a given depth, pressure acts equally on a wall, the floor, and a fish swimming there.

Pressure grows with depth and, at any one depth, pushes equally in every direction.

$p = h\rho g$ .
Deeper = higher pressure.
More dense fluid = higher pressure at same depth.
Same depth = same pressure in all directions.

Hydraulic systems

Pressure transmitted equally → small force on small piston gives large force on large piston.

Pascal's principle. Pressure applied to an enclosed fluid is transmitted UNDIMINISHED in all directions.

Hydraulic press. Two cylinders connected by fluid:

Small piston, area $A_{1}$ , force $F_{1}$ → pressure $p = F_{1}/A_{1}$ .
This pressure pushes the large piston (area $A_{2}$ ) with force $F_{2} = pA_{2} = F_{1} \times \dfrac{A_{2}}{A_{1}}$ .

Force amplification factor: $A_{2}/A_{1}$ .

Equal pressure on a larger piston gives a larger force — amplified by the area ratio A₂/A₁.

Worked. Hydraulic jack: small piston $0.001\,\text{m}^{2}$ , large piston $0.05\,\text{m}^{2}$ . Push small with $20\,\text{N}$ :

Pressure: $p = 20/0.001 = 20{,}000\,\text{Pa}$ .
Force on large piston: $20{,}000 \times 0.05 = 1000\,\text{N}$ .
Multiplied by 50.

Conservation. Energy out $\le$ energy in. The large piston moves a SMALLER distance than the small one — work done is the same (or less, with friction).

Pascal's principle: pressure transmits equally.
Hydraulic force amplification = $A_{2}/A_{1}$ .
Large piston moves smaller distance.
Used in jacks, brakes, lifts.

How it’s examined

Pressure appears most years on Paper 4 (4-6 marks) — typically a mix of $p = F/A$ for a sharp object, $p = h\rho g$ for a diver or dam, and a hydraulic-press calculation. Examiner reports flag area unit errors ( $\text{cm}^{2}$ vs $\text{m}^{2}$ ) and forgetting that hydraulics conserve energy (small input → small distance × big force).

Sources: Cambridge IGCSE Physics 0625 syllabus 2026-2028 (1.8); 0625/42 Oct/Nov 2024 — Q10 (hydraulics); 0625 Examiner Reports 2022-2024. Last reviewed 2026-05-06.

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Step-by-step worked examples — Pressure

Step-by-step solutions to past-paper-style questions on pressure, written exactly the way a tutor would explain them at the board.

1Pressure under a solid
Core• solid
Question
A box of weight $400\,\text{N}$ sits on a $0.5\,\text{m}^{2}$ base. Find the pressure on the floor.
Step-by-step solution
1. Step 1
  $p = \dfrac{F}{A}$ .
  $p = \dfrac{400}{0.5} = 800\,\text{Pa}$
Answer
$800\,\text{Pa}$
2Pressure at a depth in a liquid
Extended• Adapted from 0625/42 May/Jun 2024 Q8• fluid
Question
Find the pressure $5\,\text{m}$ below the surface of water. $\rho_{\text{water}} = 1000\,\text{kg/m}^{3}$ , $g = 10\,\text{N/kg}$ .
Step-by-step solution
1. Step 1
  $p = \rho g h$ .
  $p = 1000 \times 10 \times 5 = 50{,}000\,\text{Pa}$
Answer
$50{,}000\,\text{Pa} = 50\,\text{kPa}$
3Why a knife cuts
Core• concept
Question
Explain why a sharp knife cuts more easily than a blunt one with the same applied force.
Step-by-step solution
1. Step 1
  Sharp blade has a SMALLER contact area.
2. Step 2
  $p = F/A$ : smaller $A$ → larger pressure for the same $F$ .
Answer
Smaller contact area concentrates the same force into a higher pressure.
4Total pressure including atmosphere
Extended• atmosphere
Question
A diver $10\,\text{m}$ below the sea surface. Atmospheric pressure $= 101{,}000\,\text{Pa}$ , $\rho_{\text{seawater}} = 1030\,\text{kg/m}^{3}$ , $g = 10$ . Find the total pressure on the diver.
Step-by-step solution
1. Step 1
  Liquid contribution.
  $\Delta p = \rho g h = 1030 \times 10 \times 10 = 103{,}000\,\text{Pa}$
2. Step 2
  Add atmosphere.
  $p = 101{,}000 + 103{,}000 = 204{,}000\,\text{Pa}$
Answer
$204{,}000\,\text{Pa} = 204\,\text{kPa}$
5Pressure with $\text{cm}^{2}$ area — converting units
Core• Adapted from 0625/22 Oct/Nov 2023 Q9• units
Question
A small foot of a stool has a contact area of $25\,\text{cm}^{2}$ and supports a weight of $50\,\text{N}$ . Find the pressure in pascals.
Step-by-step solution
1. Step 1
  Convert area to m². $1\,\text{m}^{2} = 10{,}000\,\text{cm}^{2}$ .
  $A = 25 \times 10^{-4} = 2.5 \times 10^{-3}\,\text{m}^{2}$
2. Step 2
  $p = F/A$ .
  $p = \dfrac{50}{2.5 \times 10^{-3}} = 20{,}000\,\text{Pa}$
Answer
$20{,}000\,\text{Pa} = 20\,\text{kPa}$
Examiner tip
The examiner report flags candidates often substitute $25\,\text{cm}^{2}$ directly and get $2\,\text{N/cm}^{2}$ . Pa needs $\text{m}^{2}$ .
6Pressure at a depth in mercury
Extended• fluid, rho g h
Question
Find the pressure $0.76\,\text{m}$ below the surface of mercury. $\rho_{\text{Hg}} = 13{,}600\,\text{kg/m}^{3}$ , $g = 10\,\text{N/kg}$ .
Step-by-step solution
1. Step 1
  $p = \rho g h$ .
  $p = 13{,}600 \times 10 \times 0.76$
2. Step 2
  Evaluate.
  $p \approx 1.0 \times 10^{5}\,\text{Pa}$
Answer
$\approx 1.0 \times 10^{5}\,\text{Pa}$ — close to standard atmospheric pressure, which is why $760\,\text{mm Hg}$ is one atmosphere.
7U-tube manometer with two liquids
Extended• U-tube
Question
A U-tube has water in both arms in equilibrium. Oil ( $\rho = 800\,\text{kg/m}^{3}$ ) is poured into the left arm and forms a column $12\,\text{cm}$ tall above the water. Find the difference $h$ in the water levels of the two arms. $\rho_{w} = 1000\,\text{kg/m}^{3}$ .
Step-by-step solution
1. Step 1
  Pressure at the lowest common point of the U is the same from both sides.
  $\rho_{\text{oil}} g h_{\text{oil}} = \rho_{w} g h_{w}$
2. Step 2
  Cancel $g$ and solve for $h_{w}$ .
  $h_{w} = \dfrac{\rho_{\text{oil}}}{\rho_{w}} h_{\text{oil}} = \dfrac{800}{1000} \times 12 = 9.6\,\text{cm}$
Answer
Water level difference $= 9.6\,\text{cm}$ (water is higher on the right).
Examiner tip
The 2024 mark scheme awards method marks for equating pressures at the same horizontal level inside the connected liquid — not at the tops of the columns.
8Mercury barometer reading
Extended• Adapted from 0625/42 May/Jun 2024• barometer, atmosphere
Question
A mercury barometer in a lab reads a column height of $74.0\,\text{cm}$ . Take $\rho_{Hg} = 13{,}600\,\text{kg/m}^{3}$ , $g = 10\,\text{N/kg}$ . Find the atmospheric pressure in pascals.
Step-by-step solution
1. Step 1
  Convert height to metres.
  $h = 0.740\,\text{m}$
2. Step 2
  Atmospheric pressure supports the column: $p = \rho g h$ .
  $p = 13{,}600 \times 10 \times 0.740$
3. Step 3
  Evaluate.
  $p \approx 1.01 \times 10^{5}\,\text{Pa}$
Answer
$\approx 1.01 \times 10^{5}\,\text{Pa}$
Examiner tip
The examiner report flags candidates often substitute $74\,\text{cm}$ directly and get $7.4$ times the correct answer. Always convert to metres for SI.
9Convert pressure between Pa, kPa and $\text{N/cm}^{2}$
Core• units, conversion
Question
Express $85\,\text{kPa}$ in pascals and in $\text{N/cm}^{2}$ .
Step-by-step solution
1. Step 1
  $1\,\text{kPa} = 1000\,\text{Pa}$ .
  $85\,\text{kPa} = 85{,}000\,\text{Pa}$
2. Step 2
  $1\,\text{N/cm}^{2} = 10{,}000\,\text{Pa}$ , so divide by $10{,}000$ .
  $85{,}000\,\text{Pa} = 8.5\,\text{N/cm}^{2}$
Answer
$85{,}000\,\text{Pa} = 8.5\,\text{N/cm}^{2}$
10A* — hydraulic press as a pressure multiplier
Challenge• hydraulic, synoptic
Question
A hydraulic lift has a small piston of area $5.0\,\text{cm}^{2}$ and a large piston of area $200\,\text{cm}^{2}$ . A force of $30\,\text{N}$ is applied to the small piston. Find (a) the pressure transmitted through the fluid and (b) the load the large piston can support.
Step-by-step solution
1. Step 1
  Pressure on small piston.
  $p = \dfrac{F_{1}}{A_{1}} = \dfrac{30}{5.0 \times 10^{-4}} = 6.0 \times 10^{4}\,\text{Pa}$
2. Step 2
  Pressure is transmitted equally throughout the fluid (Pascal).
3. Step 3
  Force on large piston.
  $F_{2} = p A_{2} = 6.0 \times 10^{4} \times 200 \times 10^{-4} = 1200\,\text{N}$
Answer
(a) $p = 6.0 \times 10^{4}\,\text{Pa}$ . (b) Load supported $= 1200\,\text{N}$ — a $40\times$ force multiplier (matching the area ratio).
Examiner tip
The 2024 mark scheme awards method marks for stating that 'pressure is transmitted equally' — the key principle behind hydraulics.

Key Formulae — Pressure

The formulae you need to memorise for pressure on the Cambridge IGCSE 0625 paper, with every variable defined in plain English and a note on when to use it.

Pressure (general)
$p = \dfrac{F}{A}$
$p$
pressure (Pa = N/m²)
$F$
force perpendicular to surface (N)
$A$
area (m²)
When to use
Solids; or pressure of any constant force on a surface.
Pressure in a fluid (depth)
$\Delta p = \rho g h$
$\rho$
fluid density (kg/m³)
$h$
vertical depth below surface (m)
When to use
Liquids and (approximately) gases at small height changes.

Key Definitions and Keywords — Pressure

Definitions to memorise and the exact keywords mark schemes credit for pressure answers — sharpened from recent examiner reports for the 2026 0625 sitting.

Pressure
Examiner keyword
Force per unit area on a surface. SI unit pascal (Pa) = N/m².
Pascal (Pa)
Examiner keyword
$1\,\text{Pa} = 1\,\text{N/m}^{2}$ .
Atmospheric pressure
Examiner keyword
Pressure exerted by the column of air above us. About $101{,}000\,\text{Pa}$ at sea level.

Common Mistakes and Misconceptions — Pressure

The traps other students keep falling into on pressure questions — taken from recent Cambridge IGCSE 0625 examiner reports and mark schemes — and how to avoid them.

✕Using $\text{cm}^{2}$ in $p = F/A$
0625/42 — recurring
Why it happens
Reading the diagram in cm.
How to avoid it
Convert area to $\text{m}^{2}$ for the answer in pascals: $1\,\text{m}^{2} = 10{,}000\,\text{cm}^{2}$ .
✕Using horizontal distance in $\rho g h$
Why it happens
Misreading the diagram.
How to avoid it
$h$ is VERTICAL depth below the free surface.
✕Mixing g/cm³ with kg/m³ in $\rho g h$
Why it happens
Different unit systems in textbook tables.
How to avoid it
Use SI: $\rho$ in kg/m³, $h$ in m, $g$ in N/kg, then $p$ comes out in Pa.
✕Forgetting atmospheric pressure when 'total' is asked
Why it happens
Computing only $\rho g h$ .
How to avoid it
Total = atmospheric + fluid contribution. Read the question carefully.

Practice questions

Exam-style questions with step-by-step worked solutions. Try one before checking the method.

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Pressure — frequently asked questions

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Pressure

Short Study Notes in the form of Slides

Detailed Notes

What you’ll learn

How it’s examined

1Pressure under a solid

2Pressure at a depth in a liquid

3Why a knife cuts

4Total pressure including atmosphere

5Pressure with cm2\text{cm}^{2}cm2 area — converting units

6Pressure at a depth in mercury

7U-tube manometer with two liquids

8Mercury barometer reading

9Convert pressure between Pa, kPa and N/cm2\text{N/cm}^{2}N/cm2

10A* — hydraulic press as a pressure multiplier

Pressure (general)

Pressure in a fluid (depth)

Pressure

Pascal (Pa)

Atmospheric pressure

✕Using cm2\text{cm}^{2}cm2 in p=F/Ap = F/Ap=F/A

✕Using horizontal distance in ρgh\rho g hρgh

✕Mixing g/cm³ with kg/m³ in ρgh\rho g hρgh

✕Forgetting atmospheric pressure when 'total' is asked

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Pressure — frequently asked questions

5Pressure with $\text{cm}^{2}$ area — converting units

9Convert pressure between Pa, kPa and $\text{N/cm}^{2}$

✕Using $\text{cm}^{2}$ in $p = F/A$

✕Using horizontal distance in $\rho g h$

✕Mixing g/cm³ with kg/m³ in $\rho g h$