What is the definition of density in the context of Cambridge International A Levels Physics?

In Cambridge International A Levels Physics, density is defined as the mass per unit volume of a substance. It is a measure of how much matter is packed into a given volume and is calculated using the formula: Density = Mass/Volume. The standard unit of density is kilograms per cubic meter (kg/m³).

How is pressure defined and calculated in Physics?

Pressure in Physics is defined as the force exerted per unit area on the surface of an object. It is calculated using the formula: Pressure = Force/Area. The standard unit of pressure is the pascal (Pa), where 1 pascal is equivalent to 1 newton per square meter (N/m²).

How do density and pressure relate to each other in fluid mechanics?

In fluid mechanics, density and pressure are closely related. The pressure in a fluid at a given depth is directly proportional to the density of the fluid. This relationship is described by the equation: Pressure = Density × Gravitational Field Strength × Height. This means that in a denser fluid, the pressure increases more rapidly with depth.

What are some practical applications of understanding density and pressure in real-world scenarios?

Understanding density and pressure is crucial in various real-world applications. For instance, in designing ships and submarines, engineers must consider the density of water to ensure buoyancy. In meteorology, atmospheric pressure is used to predict weather patterns. Additionally, in the medical field, blood pressure measurements are vital for diagnosing and monitoring health conditions.

Why is it important to study density and pressure in the Cambridge International A Levels Physics curriculum?

Studying density and pressure in the Cambridge International A Levels Physics curriculum is important because these concepts are fundamental to understanding how forces interact with matter. They are essential for analyzing the behavior of solids, liquids, and gases under various conditions, which is critical for fields such as engineering, meteorology, and environmental science. Mastery of these concepts also lays the groundwork for more advanced studies in physics and related disciplines.

Why is "Including parallel force in pressure" a common mistake in Density and pressure?

Why it happens: Forgetting 'perpendicular'. How to avoid it: Pressure uses force COMPONENT PERPENDICULAR to surface. Parallel components don't contribute. Source: 9702 Examiner Reports 2022-2024

Why is "Using total volume instead of submerged volume" a common mistake in Density and pressure?

Why it happens: Forgetting partial submersion. How to avoid it: Upthrust uses SUBMERGED volume. Fully submerged: total volume. Floating: only submerged part. Source: 9702 Examiner Reports 2022-2024

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Detailed Study Notes

Detailed notes on Forces, Density and Pressure for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Density and Pressure Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Density and pressure. Hydrostatic pressure. Upthrust and Archimedes' principle.

At a glance

$\rho = m/V$ .
$p = F/A$ (perpendicular force).
Hydrostatic $p = \rho g h$ .
Upthrust = weight of fluid displaced.
Floating: weight = upthrust.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

4.3.1 — Compute density.
4.3.2 — Compute pressure including hydrostatic.
4.3.3 — Apply Archimedes' principle.

Density

Mass per unit volume.

Density. $\rho = m/V$ . Scalar. Units kg/m³.

Common densities.

Air: ~1.2 kg/m³.
Water: 1000 kg/m³ (or 1.0 g/cm³).
Iron: ~7900 kg/m³.
Mercury: ~13600 kg/m³.

Use. Determines floating: object less dense than fluid → floats.

Cambridge tip. Convert g/cm³ to kg/m³ by multiplying by 1000.

$\rho = m/V$ .
Water: 1000 kg/m³.
Float if $\rho_{\text{object}} < \rho_{\text{fluid}}$ .

See the full worked example for density and pressure →

Pressure

Force per unit area.

Pressure. $p = F/A$ where $F$ is perpendicular force, $A$ is area.

Units. Pa = N/m². Other: bar ( $10^5$ Pa), atm (~ $1.01 \times 10^5$ Pa).

Hydrostatic pressure (due to column of fluid). $p = \rho g h$ where $h$ is depth.

Total pressure at depth. $p_{\text{total}} = p_{\text{atm}} + \rho g h$ (often atmospheric is implicit; use just $\rho g h$ for "gauge pressure").

Derivation. Column of fluid of cross-section $A$ , height $h$ : weight = $\rho A h g$ . Pressure = weight/area = $\rho g h$ .

Direction. Pressure acts EQUALLY in ALL directions at a point in a static fluid (Pascal's principle).

Cambridge tip. Specify whether atmospheric is included.

Hydrostatic pressure p = ρgh acts equally in all directions and increases linearly with depth h.

$p = F/A$ .
Hydrostatic: $\rho g h$ .
Acts in all directions.

See the full worked example for density and pressure →

Upthrust

Weight of fluid displaced.

Archimedes' principle. Upthrust on a body in a fluid equals the WEIGHT of FLUID DISPLACED.

Formula. $U = \rho_f V_{\text{sub}} g$ .

$\rho_f$ = density of FLUID (not object).
$V_{\text{sub}}$ = submerged volume.

Why. Pressure on bottom of submerged object > pressure on top (greater depth → higher $\rho g h$ ). Net upward force = upthrust.

Floating equilibrium. Weight = upthrust. Object partially submerged just enough to displace fluid weight equal to its own weight.

Sinking. If object's weight > maximum possible upthrust (= $\rho_f V g$ ), object sinks. Equivalent: $\rho_{\text{object}} > \rho_f$ .

Example. Wood block volume $10^{-3}$ m³, floats half-submerged in water.

Submerged volume = $0.5 \times 10^{-3}$ m³.
$U = 1000 \times 0.5 \times 10^{-3} \times 9.81 = 4.91$ N.
Weight = $U$ → mass $\approx 0.5$ kg → wood density 500 kg/m³.

Cambridge tip. Mark scheme rewards explicit Archimedes' statement.

$U = \rho_f V_{\text{sub}} g$ .
Use FLUID density.
Floating: weight = upthrust.

See the full worked example for density and pressure →

Quick recap

Density $\rho = m/V$ .
Pressure $p = F/A$ .
Hydrostatic $\rho g h$ .
Upthrust = weight of fluid displaced.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\rho, p, \rho g h$ formulas.
Archimedes' principle.
Floating condition.

How it’s examined

Density and pressure appear every Paper 1 and Paper 2. Most-tested: hydrostatic pressure (6 marks), upthrust calculation (8 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Density and pressure

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

1Density (4 marks)
Extended• Adapted from 9702/22 May/Jun 2024• density
Question
A block has mass $0.50$ kg and volume $2.0 \times 10^{-4}$ m³. Find its density. (4 marks)
Step-by-step solution
1. Step 1
  $\rho = m/V$ .
  $\rho = \dfrac{0.50}{2.0 \times 10^{-4}} = 2500 \text{ kg/m}^3$
Answer
$\rho = 2500$ kg/m³.
2Hydrostatic pressure (6 marks)
Extended• hydrostatic pressure
Question
Find the pressure due to water at the bottom of a swimming pool 2.5 m deep. $\rho = 1000$ kg/m³, $g = 9.81$ . (6 marks)
Step-by-step solution
1. Step 1
  $p = \rho g h$ .
  $p = 1000 \times 9.81 \times 2.5 = 24525 \text{ Pa}$
2. Step 2
  Round. $p \approx 2.45 \times 10^4$ Pa or 24.5 kPa.
Answer
$p \approx 2.45 \times 10^4$ Pa.
3Archimedes' principle (8 marks)
Extended• upthrust
Question
A wooden block of volume $1.0 \times 10^{-3}$ m³ floats half-submerged in water. Find (a) upthrust, (b) weight of block, (c) density of wood. Take $\rho_w = 1000$ kg/m³, $g = 9.81$ . (8 marks)
Step-by-step solution
1. Step 1
  Volume submerged = half of total = $0.5 \times 10^{-3}$ m³.
2. Step 2
  Upthrust = weight of displaced water.
  $U = \rho_w V_{\text{sub}} g = 1000 \times 0.5 \times 10^{-3} \times 9.81 \approx 4.91 \text{ N}$
3. Step 3
  Floating equilibrium: weight = upthrust.
  $W = 4.91 \text{ N}$
4. Step 4
  Mass of block.
  $m = W/g = 4.91/9.81 \approx 0.50 \text{ kg}$
5. Step 5
  Density of wood.
  $\rho_{\text{wood}} = m/V = 0.50/(1.0 \times 10^{-3}) = 500 \text{ kg/m}^3$
Answer
(a) $U \approx 4.91$ N. (b) $W \approx 4.91$ N. (c) $\rho_{\text{wood}} = 500$ kg/m³.

Key Formulae — Density and pressure

The formulae you need to memorise for density and pressure on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Density
$\rho = \dfrac{m}{V}$
$\rho$
density (kg/m³)
When to use
Mass per unit volume. Units: kg/m³.
Pressure
$p = \dfrac{F}{A}$
$F$
force perpendicular to surface
$A$
area
When to use
Force per unit area. Units: Pa = N/m².
Hydrostatic pressure
$p = \rho g h$
$h$
depth below surface
When to use
Pressure due to a column of fluid. Excludes atmospheric pressure; for total, add $p_{\text{atm}}$ .
Upthrust (Archimedes)
$U = \rho_f V_{\text{sub}} g$
$\rho_f$
fluid density
$V_{\text{sub}}$
submerged volume
When to use
Buoyant force on an immersed/floating body. Equals weight of fluid displaced.

Key Definitions and Keywords — Density and pressure

Definitions to memorise and the exact keywords mark schemes credit for density and pressure answers — sharpened from recent examiner reports for the 2026 Cambridge International A Level 9702 sitting.

Density
Examiner keyword
Mass per unit volume. $\rho = m/V$ . Units: kg/m³.
Pressure
Examiner keyword
Force per unit area, acting perpendicular to surface. $p = F/A$ . Units: Pa = N/m².
Upthrust
Examiner keyword
Upward force on a body immersed in a fluid. Equals weight of fluid displaced.
Archimedes' principle
Examiner keyword
Upthrust on a body immersed in a fluid equals the weight of fluid displaced.

Common Mistakes and Misconceptions — Density and pressure

The traps other students keep falling into on density and pressure questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Including parallel force in pressure
9702 Examiner Reports 2022-2024
Why it happens
Forgetting 'perpendicular'.
How to avoid it
Pressure uses force COMPONENT PERPENDICULAR to surface. Parallel components don't contribute.
✕Using total volume instead of submerged volume
9702 Examiner Reports 2022-2024
Why it happens
Forgetting partial submersion.
How to avoid it
Upthrust uses SUBMERGED volume. Fully submerged: total volume. Floating: only submerged part.

Density and pressure — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Cambridge International A Levels Physics (9702)

Density and pressure

0% Complete

3 tasks remaining

Short Notes - Density and pressure

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Detailed Study Notes

Detailed notes on Forces, Density and Pressure for Cambridge International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Density and Pressure Study Notes — Cambridge International A Level Physics 9702 (2025-2027 syllabus)

Density and pressure. Hydrostatic pressure. Upthrust and Archimedes' principle.

At a glance

$\rho = m/V$ .
$p = F/A$ (perpendicular force).
Hydrostatic $p = \rho g h$ .
Upthrust = weight of fluid displaced.
Floating: weight = upthrust.

What you’ll learn

Mapped to the Cambridge International A Level 9702 syllabus (2025-2027).

4.3.1 — Compute density.
4.3.2 — Compute pressure including hydrostatic.
4.3.3 — Apply Archimedes' principle.

Density

Mass per unit volume.

Density. $\rho = m/V$ . Scalar. Units kg/m³.

Common densities.

Air: ~1.2 kg/m³.
Water: 1000 kg/m³ (or 1.0 g/cm³).
Iron: ~7900 kg/m³.
Mercury: ~13600 kg/m³.

Use. Determines floating: object less dense than fluid → floats.

Cambridge tip. Convert g/cm³ to kg/m³ by multiplying by 1000.

$\rho = m/V$ .
Water: 1000 kg/m³.
Float if $\rho_{\text{object}} < \rho_{\text{fluid}}$ .

See the full worked example for density and pressure →

Pressure

Force per unit area.

Pressure. $p = F/A$ where $F$ is perpendicular force, $A$ is area.

Units. Pa = N/m². Other: bar ( $10^5$ Pa), atm (~ $1.01 \times 10^5$ Pa).

Hydrostatic pressure (due to column of fluid). $p = \rho g h$ where $h$ is depth.

Total pressure at depth. $p_{\text{total}} = p_{\text{atm}} + \rho g h$ (often atmospheric is implicit; use just $\rho g h$ for "gauge pressure").

Derivation. Column of fluid of cross-section $A$ , height $h$ : weight = $\rho A h g$ . Pressure = weight/area = $\rho g h$ .

Direction. Pressure acts EQUALLY in ALL directions at a point in a static fluid (Pascal's principle).

Cambridge tip. Specify whether atmospheric is included.

Hydrostatic pressure p = ρgh acts equally in all directions and increases linearly with depth h.

$p = F/A$ .
Hydrostatic: $\rho g h$ .
Acts in all directions.

See the full worked example for density and pressure →

Upthrust

Weight of fluid displaced.

Archimedes' principle. Upthrust on a body in a fluid equals the WEIGHT of FLUID DISPLACED.

Formula. $U = \rho_f V_{\text{sub}} g$ .

$\rho_f$ = density of FLUID (not object).
$V_{\text{sub}}$ = submerged volume.

Why. Pressure on bottom of submerged object > pressure on top (greater depth → higher $\rho g h$ ). Net upward force = upthrust.

Floating equilibrium. Weight = upthrust. Object partially submerged just enough to displace fluid weight equal to its own weight.

Sinking. If object's weight > maximum possible upthrust (= $\rho_f V g$ ), object sinks. Equivalent: $\rho_{\text{object}} > \rho_f$ .

Example. Wood block volume $10^{-3}$ m³, floats half-submerged in water.

Submerged volume = $0.5 \times 10^{-3}$ m³.
$U = 1000 \times 0.5 \times 10^{-3} \times 9.81 = 4.91$ N.
Weight = $U$ → mass $\approx 0.5$ kg → wood density 500 kg/m³.

Cambridge tip. Mark scheme rewards explicit Archimedes' statement.

$U = \rho_f V_{\text{sub}} g$ .
Use FLUID density.
Floating: weight = upthrust.

See the full worked example for density and pressure →

Quick recap

Density $\rho = m/V$ .
Pressure $p = F/A$ .
Hydrostatic $\rho g h$ .
Upthrust = weight of fluid displaced.

Memorise this

Verbatim phrases and definitions Cambridge mark schemes credit.

$\rho, p, \rho g h$ formulas.
Archimedes' principle.
Floating condition.

How it’s examined

Density and pressure appear every Paper 1 and Paper 2. Most-tested: hydrostatic pressure (6 marks), upthrust calculation (8 marks).

Sources: Cambridge International A Level Physics 9702 syllabus (2025-2027); 9702 Examiner Reports 2022-2024; 9702/22 May/Jun 2024 question paper and mark scheme. Last reviewed 2026-05-11.

Take this whole topic with you

Step-by-step worked examples — Density and pressure

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

1Density (4 marks)
Extended• Adapted from 9702/22 May/Jun 2024• density
Question
A block has mass $0.50$ kg and volume $2.0 \times 10^{-4}$ m³. Find its density. (4 marks)
Step-by-step solution
1. Step 1
  $\rho = m/V$ .
  $\rho = \dfrac{0.50}{2.0 \times 10^{-4}} = 2500 \text{ kg/m}^3$
Answer
$\rho = 2500$ kg/m³.
2Hydrostatic pressure (6 marks)
Extended• hydrostatic pressure
Question
Find the pressure due to water at the bottom of a swimming pool 2.5 m deep. $\rho = 1000$ kg/m³, $g = 9.81$ . (6 marks)
Step-by-step solution
1. Step 1
  $p = \rho g h$ .
  $p = 1000 \times 9.81 \times 2.5 = 24525 \text{ Pa}$
2. Step 2
  Round. $p \approx 2.45 \times 10^4$ Pa or 24.5 kPa.
Answer
$p \approx 2.45 \times 10^4$ Pa.
3Archimedes' principle (8 marks)
Extended• upthrust
Question
A wooden block of volume $1.0 \times 10^{-3}$ m³ floats half-submerged in water. Find (a) upthrust, (b) weight of block, (c) density of wood. Take $\rho_w = 1000$ kg/m³, $g = 9.81$ . (8 marks)
Step-by-step solution
1. Step 1
  Volume submerged = half of total = $0.5 \times 10^{-3}$ m³.
2. Step 2
  Upthrust = weight of displaced water.
  $U = \rho_w V_{\text{sub}} g = 1000 \times 0.5 \times 10^{-3} \times 9.81 \approx 4.91 \text{ N}$
3. Step 3
  Floating equilibrium: weight = upthrust.
  $W = 4.91 \text{ N}$
4. Step 4
  Mass of block.
  $m = W/g = 4.91/9.81 \approx 0.50 \text{ kg}$
5. Step 5
  Density of wood.
  $\rho_{\text{wood}} = m/V = 0.50/(1.0 \times 10^{-3}) = 500 \text{ kg/m}^3$
Answer
(a) $U \approx 4.91$ N. (b) $W \approx 4.91$ N. (c) $\rho_{\text{wood}} = 500$ kg/m³.

Key Formulae — Density and pressure

The formulae you need to memorise for density and pressure on the Cambridge International A Level 9702 paper, with every variable defined in plain English and a note on when to use it.

Density
$\rho = \dfrac{m}{V}$
$\rho$
density (kg/m³)
When to use
Mass per unit volume. Units: kg/m³.
Pressure
$p = \dfrac{F}{A}$
$F$
force perpendicular to surface
$A$
area
When to use
Force per unit area. Units: Pa = N/m².
Hydrostatic pressure
$p = \rho g h$
$h$
depth below surface
When to use
Pressure due to a column of fluid. Excludes atmospheric pressure; for total, add $p_{\text{atm}}$ .
Upthrust (Archimedes)
$U = \rho_f V_{\text{sub}} g$
$\rho_f$
fluid density
$V_{\text{sub}}$
submerged volume
When to use
Buoyant force on an immersed/floating body. Equals weight of fluid displaced.

Key Definitions and Keywords — Density and pressure

Density
Examiner keyword
Mass per unit volume. $\rho = m/V$ . Units: kg/m³.
Pressure
Examiner keyword
Force per unit area, acting perpendicular to surface. $p = F/A$ . Units: Pa = N/m².
Upthrust
Examiner keyword
Upward force on a body immersed in a fluid. Equals weight of fluid displaced.
Archimedes' principle
Examiner keyword
Upthrust on a body immersed in a fluid equals the weight of fluid displaced.

Common Mistakes and Misconceptions — Density and pressure

The traps other students keep falling into on density and pressure questions — taken from recent Cambridge International A Level 9702 examiner reports and mark schemes — and how to avoid them.

✕Including parallel force in pressure
9702 Examiner Reports 2022-2024
Why it happens
Forgetting 'perpendicular'.
How to avoid it
Pressure uses force COMPONENT PERPENDICULAR to surface. Parallel components don't contribute.
✕Using total volume instead of submerged volume
9702 Examiner Reports 2022-2024
Why it happens
Forgetting partial submersion.
How to avoid it
Upthrust uses SUBMERGED volume. Fully submerged: total volume. Floating: only submerged part.

Density and pressure — frequently asked questions

The things students keep getting wrong in this sub-topic, answered.

Density and pressure

Short Notes - Density and pressure

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Density (4 marks)

2Hydrostatic pressure (6 marks)

3Archimedes' principle (8 marks)

Density

Pressure

Hydrostatic pressure

Upthrust (Archimedes)

Density

Pressure

Upthrust

Archimedes' principle

✕Including parallel force in pressure

✕Using total volume instead of submerged volume

Density and pressure — frequently asked questions

Density and pressure

Short Notes - Density and pressure

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Density (4 marks)

2Hydrostatic pressure (6 marks)

3Archimedes' principle (8 marks)

Density

Pressure

Hydrostatic pressure

Upthrust (Archimedes)

Density

Pressure

Upthrust

Archimedes' principle

✕Including parallel force in pressure

✕Using total volume instead of submerged volume

Density and pressure — frequently asked questions