What is the formula for calculating density in Physics?

In Physics, particularly in the Edexcel IGCSE curriculum, density is calculated using the formula: Density = Mass / Volume. This formula helps determine how much mass is contained in a given volume of a substance, whether it be a solid, liquid, or gas.

How does pressure differ in solids, liquids, and gases?

Pressure in solids is typically exerted in a specific direction, often calculated as force per unit area. In liquids, pressure is exerted equally in all directions and increases with depth due to the weight of the liquid above. In gases, pressure is also exerted equally in all directions and is influenced by temperature and volume, as described by the ideal gas law.

Why does an object float or sink in a liquid?

An object floats or sinks in a liquid based on its density compared to the liquid's density. If the object's density is less than the liquid's density, it will float. Conversely, if the object's density is greater, it will sink. This principle is crucial in understanding buoyancy and is part of the Edexcel IGCSE Physics curriculum.

How is pressure calculated in a liquid at a certain depth?

Pressure in a liquid at a certain depth is calculated using the formula: Pressure = Density of liquid × Gravitational field strength × Depth. This formula shows that pressure increases with depth due to the weight of the liquid above, a key concept in the study of fluids in Physics.

What are some real-life applications of understanding density and pressure?

Understanding density and pressure has numerous real-life applications, such as designing ships and submarines, predicting weather patterns, and creating hydraulic systems. These concepts are integral to various fields, including engineering, meteorology, and medicine, and are covered in the Edexcel IGCSE Physics syllabus.

Why is "Mixing g/cm³ and kg/m³ in a single calculation" a common mistake in Density and Pressure?

Why it happens: Question gives mass in grams and volume in cm³ but asks for density in kg/m³. How to avoid it: Convert at the START: 1 g = 10⁻³ kg; 1 cm³ = 10⁻⁶ m³. Or do the calculation in g/cm³ and × 1000 at the end. Source: 4PH1 Examiner Reports 2022-2024

Why is "Quoting an area in cm² in the pressure formula" a common mistake in Density and Pressure?

Why it happens: Diagrams give dimensions in cm. How to avoid it: p in pascals REQUIRES area in m². Convert: 1 cm² = 10⁻⁴ m². 100 N over 50 cm² = 100 / 0.005 m² = 20 000 Pa. Source: 4PH1 Examiner Reports 2022-2024

Why is "Saying $p = h\rho g$ gives the TOTAL pressure (forgetting atmospheric)" a common mistake in Density and Pressure?

Why it happens: Misreading the formula's meaning. How to avoid it: Spec 5.7 formula gives the EXCESS pressure due to the FLUID COLUMN above. Total pressure on a submerged object = p_atm + h g. Read the question carefully.

Why is "Saying pressure 'pushes downwards' in a static fluid" a common mistake in Density and Pressure?

Why it happens: Everyday image of weight pushing on the floor. How to avoid it: Spec 5.6: pressure in a fluid at rest acts EQUALLY IN ALL DIRECTIONS at any point. The horizontal sides of a tank experience pressure too — that's why dams are built thicker at the base.

Why is "Confusing the volume of a regular solid with a circle/area formula" a common mistake in Density and Pressure?

Why it happens: Memorising formulas without practising. How to avoid it: Cuboid: V = lwh. Cylinder: V = r^2 h. Sphere: V = 43 r^3. For irregular shapes, use water displacement (spec 5.4). Source: 4PH1/1P Examiner Reports 2022-2024

Solids, Liquids and GasesPearson Edexcel IGCSE Physics 4PH1 Higher

Density and Pressure

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Density and Pressure — Pearson Edexcel International GCSE Physics 4PH1 Study Notes (2026 onwards, Spec Issue 4)

Density $\rho = m/V$ with the required practical for regular/irregular solids and liquids, pressure $p = F/A$ , pressure at a point in a fluid at rest acts equally in all directions, and pressure difference at depth $p = h\rho g$ .

What you’ll learn

Mapped to the Pearson Edexcel IGCSE 4PH1 syllabus (2026 onwards).

5.1 — Use the units °C, K, J, kg, kg/m³, m, m², m³, m/s, m/s², N and Pa.
5.2P — Use the unit J/(kg °C) for specific heat capacity (Paper 2).
5.3 — Know and use the relationship: density = mass / volume, $\rho = m/V$ .
5.4 — Required practical: investigate density using direct measurements of mass and volume (for regular solids, irregular solids and liquids).
5.5 — Know and use the relationship: pressure = force / area, $p = F/A$ .
5.6 — Understand that the pressure at a point in a gas or liquid which is at rest acts equally in all directions.
5.7 — Know and use the relationship: pressure difference = height × density × g, $p = h\rho g$ .

Units in this topic (spec 5.1, 5.2P)

Temperature, density, pressure units.

Edexcel 4PH1 awards marks for the correct unit on every numerical answer. For Topic 5:

Temperature in °C or K (kelvin); kelvin scale used for ideal-gas calculations (spec 5.17).
Energy in joules (J).
Mass in kg.
Density in kg/m³ (NOT g/cm³ for SI answers — but 1 g/cm³ = 1000 kg/m³).
Length in m; area in m²; volume in m³.
Speed in m/s; acceleration in m/s².
Force in N; pressure in pascals (Pa = N/m²).

Paper 2 only (spec 5.2P). Specific heat capacity has unit J/(kg °C) — joules per kilogram per degree Celsius. Used in spec 5.13P calculations.

Density: kg/m³ in SI (1 g/cm³ = 1000 kg/m³).
Pressure: pascal (Pa) = N/m².
Specific heat capacity (Paper 2 only): J/(kg °C).
Always state the unit alongside the final answer.

Density (spec 5.3)

Mass per unit volume.

Definition. The MASS of a substance contained in unit VOLUME.

$\rho = \dfrac{m}{V}$

$\rho$ in kg/m³ (or g/cm³)
$m$ in kg (or g)
$V$ in m³ (or cm³)

Key values to know.

Water: 1000 kg/m³ (= 1.0 g/cm³). Define by historical convention.
Air at room temperature: ≈ 1.2 kg/m³ (much less than water — that's why air rises in water).
Ice: 920 kg/m³ — less than water, so ice floats.
Aluminium: 2700 kg/m³.
Steel: 7800 kg/m³.
Lead: 11 300 kg/m³.

Why density matters. Floats and sinks: an object less dense than the fluid it's in will FLOAT; one denser will SINK. Hot air rises in cooler air (lower density when hot, spec 4.6 — convection).

Volume of regular solids.

Cuboid: $V = lwh$ .
Cylinder: $V = \pi r^2 h$ .
Sphere: $V = \tfrac{4}{3}\pi r^3$ .

Volume of liquid. Read from a measuring cylinder — at eye level, at the BOTTOM of the meniscus.

Volume of irregular solid. Water displacement: $V = V_{2} - V_{1}$ (spec 5.4 required practical).

$\rho = m/V$ — kg/m³.
1 g/cm³ = 1000 kg/m³.
Water = 1000 kg/m³; air = 1.2 kg/m³.
Regular solid: measure dimensions. Irregular: water displacement.

See the full worked example for density and pressure →

Required practical: density investigations (spec 5.4)

Three sub-techniques.

Spec 5.4 lists THREE situations for measuring density. Memorise all three methods.

(1) Regular solid (e.g. a metal cuboid).

Use a balance to measure mass $m$ .
Use a ruler (or vernier callipers for accuracy) to measure dimensions.
Calculate volume from the appropriate formula ( $lwh$ for cuboid, $\pi r^2 h$ for cylinder).
$\rho = m/V$ .

(2) Irregular solid (e.g. a stone).

Use a balance to measure mass $m$ .
Half-fill a measuring cylinder with water. Read initial volume $V_1$ at eye level.
Gently submerge the object using a thread (no splashing!).
Read final volume $V_2$ .
Volume of object = $V_2 - V_1$ .
$\rho = m/(V_2 - V_1)$ .

(3) Liquid.

Find the mass $m_1$ of an EMPTY measuring cylinder.
Pour in a known volume of liquid $V$ .
Find the new mass $m_2$ .
Mass of liquid = $m_2 - m_1$ .
$\rho = (m_2 - m_1) / V$ .

Sources of error.

Parallax when reading the meniscus → eye level matters.
Air bubbles clinging to an irregular solid → over-estimate volume → under-estimate density.
For a regular solid, ruler resolution (0.5 mm) can be significant on small objects.

Improvement. Repeat each measurement at least 3 times and take the mean. For higher accuracy, use vernier callipers for solid dimensions and an electronic balance reading to 0.01 g.

The rise in water level equals the volume of the submerged irregular solid.

Regular solid: measure dimensions; $V = lwh$ or $\pi r^2 h$ .
Irregular solid: water displacement; $V = V_2 - V_1$ .
Liquid: mass difference (full − empty cylinder) ÷ volume.
Repeat × 3 to reduce random error.

See the full worked example for density and pressure →

Pressure (spec 5.5)

Force per unit area.

Definition. Pressure = force applied PERPENDICULAR to a surface, divided by the area of that surface.

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

$p$ in pascals (Pa = N/m²)
$F$ in newtons (N) — force component normal to the surface
$A$ in square metres (m²)

Intuitive picture. The same force spread over a LARGER area gives LOWER pressure.

Walking on snow with snowshoes: large area → low pressure → don't sink in.
High-heel shoes: tiny area → enormous pressure (~MPa) → can dent floors.
Drawing pin: blunt end vs sharp end concentrates force on a tiny area at the tip.
Tractor tyres: large area → low pressure → don't compact soil.
Knives: sharp edge has small contact area → high pressure → cuts easily.

1 Pa = 1 N/m². Small unit — atmospheric pressure is ~10⁵ Pa, and tyre pressure ~200 kPa.

Pressure also applies to fluids. A gas in a sealed container pushes on the walls. The force per unit area on each wall is the gas pressure.

$p = F/A$ .
Pa = N/m² — small unit.
Smaller area at same force → higher pressure.
Atmospheric pressure ≈ 10⁵ Pa.

See the full worked example for density and pressure →

Pressure in static fluids — direction (spec 5.6)

Equally in all directions.

Spec 5.6 statement. The pressure at a point in a gas or liquid which is at rest acts EQUALLY in all directions.

What this means. Imagine a tiny spherical balloon submerged in water. Water molecules hit every part of the balloon's surface. At the same DEPTH, the force per unit area (pressure) is the SAME no matter which direction you choose to measure it.

Practical examples.

A diving bell with windows on all sides: water pressure pushes inwards on every window equally (at the same depth).
A balloon underwater: squeezed uniformly from all sides.
A submerged object experiences a NET UPWARD force (buoyancy = upthrust) because pressure increases with depth — the bottom is at higher pressure than the top — but at any one depth the pressure is direction-independent.

Why dams are thicker at the base. Pressure increases with depth (spec 5.7), so the bottom of the wall holds back more pressure. Pressure pushes HORIZONTALLY on the wall at depth — so the wall must be strong horizontally, not just vertically.

At a point, pressure is the same in every direction.
Pressure still varies WITH DEPTH (spec 5.7).
Why dams are thicker at the base.

See the full worked example for density and pressure →

Pressure at depth (spec 5.7)

$p = h\rho g$ — linear with depth.

Formula. The pressure DIFFERENCE between a point at depth $h$ and the surface of a static fluid of density $\rho$ :

$p = h \rho g$

$p$ in pascals (Pa)
$h$ = VERTICAL depth (m)
$\rho$ = density of the fluid (kg/m³)
$g$ = 9.8 N/kg (often 10 in 4PH1 questions)

Derivation (not required but illustrative). A column of fluid of cross-sectional area $A$ , height $h$ , has volume $V = A h$ , mass $m = \rho V = \rho A h$ , weight $W = mg = \rho A h g$ . The force is spread over area $A$ , so pressure = $W/A = h \rho g$ .

Worked examples.

10 m of water: $p = 10 \times 1000 \times 10 = 10^5$ Pa. So every 10 m of depth adds ~1 atmosphere of pressure. A diver at 30 m experiences ~3 atmospheres of water pressure (+ 1 atmosphere of air → total 4 atmospheres).
The Mariana Trench is ~11 km deep — pressure ~1100 atmospheres (over 100 MPa).

Key features of $p = h\rho g$ .

Pressure depends ONLY on depth $h$ , density $\rho$ and $g$ — NOT on the shape of the container.
Two different-shaped vessels filled with water to the same height have the same pressure at the bottom (the 'hydrostatic paradox').
Pressure due to atmosphere (~ $10^{5}$ Pa) adds on top if you want TOTAL pressure.

Application: U-tube manometer. A U-shaped tube with liquid in it. Connect one side to a gas → the height difference $h$ between the two columns gives the gas pressure relative to atmospheric: $p_{\text{gas}} - p_{\text{atm}} = h\rho g$ .

Pressure grows with depth — the deepest hole has the highest pressure and the furthest jet.

$p = h\rho g$ — pressure DIFFERENCE due to height of fluid.
Linear: 10 m water ≈ 1 atmosphere extra.
Independent of container shape.
Total pressure on submerged object = atmospheric + $h\rho g$ .

See the full worked example for density and pressure →

How it’s examined

Density and pressure appears on every 4PH1 paper — usually 6-10 marks across Paper 1 and Paper 2. Highest-frequency items: density calculations with unit conversions (3-4 marks); the required practical 5.4 method (5-6 marks); pressure $p = F/A$ short calculations (3-4 marks); pressure-depth $p = h\rho g$ calculations (3-4 marks). Spec 5.6 (pressure direction) is a 1-2 mark add-on in extended-response questions.

Sources: Pearson Edexcel International GCSE Physics 4PH1 specification — Issue 4, September 2024 (valid for 2026 onwards); Pearson Edexcel 4PH1 Examiner Reports 2022-2024; 4PH1/1P May/Jun 2024 question paper and mark scheme; 4PH1/1P January 2024 question paper and mark scheme. Last reviewed 2026-05-12.

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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 of a block (3 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• density, spec-5.3, Paper 1
Question
A rectangular block measures 8.0 cm × 5.0 cm × 2.0 cm and has a mass of 320 g. Calculate the density of the block in kg/m³. (3 marks)
Step-by-step solution
1. Step 1
  Convert to SI units. Mass: 320 g = 0.320 kg. Volume: 8.0 × 5.0 × 2.0 = 80 cm³ = 80 × 10⁻⁶ m³ = 8.0 × 10⁻⁵ m³.
2. Step 2
  Write the relationship (spec 5.3).
  $\rho = \dfrac{m}{V}$
3. Step 3
  Substitute and evaluate.
  $\rho = \dfrac{0.320}{8.0 \times 10^{-5}} = 4000\ \text{kg/m}^3$
Answer
Density = $4.0 \times 10^{3}$ kg/m³.
Examiner tip
Edexcel mark scheme awards: 1 mark unit conversion, 1 mark formula/substitution, 1 mark final answer with unit. Many candidates leave answers in g/cm³ — 4 g/cm³ is the same density but you must follow the requested unit. To convert: 1 g/cm³ = 1000 kg/m³.
2Required practical: density of an irregular solid (6 marks, Paper 1)
Core• Adapted from 4PH1/1P January 2024• required practical, spec-5.4, Paper 1
Question
Describe a method to find the density of an irregularly-shaped stone using only a balance, a measuring cylinder, water, and a piece of thread. (6 marks)
Step-by-step solution
1. Step 1
  Measure mass (1 mark). Use a balance to find the mass $m$ of the stone in kilograms (or grams, then convert).
2. Step 2
  Initial water volume (1 mark). Half-fill a measuring cylinder with water. Read and record the volume $V_1$ (read at the bottom of the meniscus, at eye-level).
3. Step 3
  Submerge the stone (1 mark). Tie thread to the stone and gently lower it until fully submerged, taking care not to splash water out.
4. Step 4
  Final volume (1 mark). Read the new volume $V_2$ . The stone's volume = $V_2 − V_1$ . Convert cm³ to m³ (×10⁻⁶).
5. Step 5
  Calculate density (1 mark).
  $\rho = \dfrac{m}{V_2 - V_1}$
6. Step 6
  Reduce random error (1 mark). Repeat 3 times; take the mean of the volume readings. Ensure no air bubbles cling to the stone (would over-estimate volume → under-estimate density).
Answer
Mass on balance; measure displaced water volume in cylinder (water-displacement method); $\rho = m/(V_2 - V_1)$ . Repeat for accuracy.
Examiner tip
Spec 5.4 is a REQUIRED PRACTICAL. Edexcel mark schemes reward: balance for mass, displacement for irregular volume, calculation, AND attention to errors (bubbles, parallax in reading the meniscus). For a regular solid, MEASURE the dimensions instead — $V = lwh$ (rectangle) or $V = \pi r^2 h$ (cylinder).
3Pressure under feet (3 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• pressure, spec-5.5, Paper 1
Question
A person of weight 600 N stands on the ground. The total area of contact between her feet and the ground is 0.030 m². Calculate the pressure exerted on the ground. (3 marks)
Step-by-step solution
1. Step 1
  Write the relationship (spec 5.5).
  $p = \dfrac{F}{A}$
2. Step 2
  Substitute.
  $p = \dfrac{600}{0.030}$
3. Step 3
  Evaluate and quote the unit.
  $p = 20\,000\ \text{Pa} = 2.0 \times 10^{4}\ \text{Pa}$
Answer
Pressure = $2.0 \times 10^{4}$ Pa (20 kPa).
Examiner tip
Edexcel: 1 mark formula, 1 substitution, 1 final answer with unit (Pa). The unit pascal = N/m². If only ONE foot is on the ground (one-legged stand), pressure DOUBLES because area halves. Heels of stiletto shoes exert pressures of MPa — large enough to dent floors.
4Pressure at depth in a lake (4 marks, Paper 1)
Extended• Adapted from 4PH1/1P January 2024• pressure difference, depth, spec-5.7, Paper 1
Question
A diver descends to a depth of 15 m below the surface of a freshwater lake. Density of fresh water = 1000 kg/m³; $g = 10$ N/kg. Calculate the EXCESS pressure (above atmospheric) that the diver experiences from the water alone. (4 marks)
Step-by-step solution
1. Step 1
  Write the relationship (spec 5.7).
  $p = h \rho g$
2. Step 2
  Substitute.
  $p = 15 \times 1000 \times 10$
3. Step 3
  Evaluate.
  $p = 150\,000\ \text{Pa} = 1.5 \times 10^{5}\ \text{Pa}$
4. Step 4
  Context. Atmospheric pressure at sea level is also $\approx 1 \times 10^{5}$ Pa, so the TOTAL pressure on the diver = atmospheric + water = $2.5 \times 10^{5}$ Pa (about 2.5 times atmospheric).
Answer
Excess pressure = $1.5 \times 10^{5}$ Pa (= 150 kPa).
Examiner tip
Spec 5.7 formula gives the pressure DIFFERENCE between two depths (or between depth $h$ and the surface). To get TOTAL pressure on a diver, ADD atmospheric pressure (~10⁵ Pa). Pressure increases linearly with depth — useful for dam-thickness questions.
5Pressure in a static fluid acts in all directions (3 marks, Paper 1)
Core• Adapted from 4PH1/1P May/Jun 2024• pressure direction, spec-5.6, Paper 1
Question
A small balloon at the bottom of a swimming pool is squeezed equally from all sides. Explain this observation in terms of pressure in a liquid at rest. (3 marks)
Step-by-step solution
1. Step 1
  Statement (spec 5.6). In a liquid (or gas) at rest, the pressure at a point acts EQUALLY in ALL directions.
2. Step 2
  Application. Water molecules surround the balloon. Each face of the balloon experiences the same water pressure pushing INWARD because pressure at that depth is the same in every direction.
3. Step 3
  Implication. The balloon is compressed symmetrically — it does NOT bulge outwards on the bottom because the bottom of the water doesn't push more strongly sideways than the rest.
Answer
Pressure in a fluid at rest acts EQUALLY in all directions at any one point → balloon squeezed evenly.
Examiner tip
Spec 5.6 statement is short but recurrent. Edexcel rewards: (1) at rest, (2) at a point, (3) equally in all directions. The pressure does INCREASE with depth (spec 5.7), so the bottom of the balloon is at slightly higher pressure than the top — but at any one point pressure is the same in all directions.

Key Formulae — Density and Pressure

The formulae you need to memorise for density and pressure on the Pearson Edexcel IGCSE 4PH1 paper, with every variable defined in plain English and a note on when to use it.

Density (spec 5.3)
$\rho = \dfrac{m}{V}$
When to use
$\rho$ = density (kg/m³), $m$ = mass (kg), $V$ = volume (m³). Common conversion: 1 g/cm³ = 1000 kg/m³. Water density ≈ 1000 kg/m³; air ≈ 1.2 kg/m³; steel ≈ 7800 kg/m³.
Pressure (spec 5.5)
$p = \dfrac{F}{A}$
When to use
$p$ in pascals (Pa = N/m²), $F$ = perpendicular force on the surface (N), $A$ = area of surface (m²). Use for solids in contact with surfaces, gases against container walls, and any straightforward 'force per area' calculation.
Pressure difference in a fluid (spec 5.7)
$p = h \rho g$
When to use
$p$ = pressure difference (Pa), $h$ = vertical depth/height difference (m), $\rho$ = density of fluid (kg/m³), $g$ = 9.8 N/kg (or 10 in 4PH1 questions). Gives pressure due to the WEIGHT of a fluid column. Add atmospheric pressure if you want TOTAL pressure.

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 Pearson Edexcel IGCSE 4PH1 sitting.

Density (spec 5.3)
Examiner keyword
The mass per unit volume of a substance. $\rho = m/V$ . SI unit: kg/m³ (often quoted as g/cm³ — convert by × 1000).
Pressure (spec 5.5)
Examiner keyword
The force per unit area applied perpendicular to a surface. $p = F/A$ . SI unit: pascal (Pa) = newton per square metre (N/m²). 1 atmospheric pressure ≈ 1.0 × 10⁵ Pa.
Pressure direction in a static fluid (spec 5.6)
Examiner keyword
At any single point inside a liquid or gas at rest, the pressure acts EQUALLY in all directions. The pressure on a small flat surface depends on its depth, not its orientation.
Pressure depth relationship (spec 5.7)
Examiner keyword
The pressure difference $\Delta p$ between two points separated by a vertical depth $h$ in a fluid of density $\rho$ is $\Delta p = h \rho g$ . Pressure INCREASES with depth.

Common Mistakes and Misconceptions — Density and Pressure

The traps other students keep falling into on density and pressure questions — taken from recent Pearson Edexcel IGCSE 4PH1 examiner reports and mark schemes — and how to avoid them.

✕Mixing g/cm³ and kg/m³ in a single calculation
4PH1 Examiner Reports 2022-2024
Why it happens
Question gives mass in grams and volume in cm³ but asks for density in kg/m³.
How to avoid it
Convert at the START: 1 g = 10⁻³ kg; 1 cm³ = 10⁻⁶ m³. Or do the calculation in g/cm³ and × 1000 at the end.
✕Quoting an area in cm² in the pressure formula
4PH1 Examiner Reports 2022-2024
Why it happens
Diagrams give dimensions in cm.
How to avoid it
$p$ in pascals REQUIRES area in m². Convert: 1 cm² = 10⁻⁴ m². 100 N over 50 cm² = 100 / 0.005 m² = 20 000 Pa.
✕Saying $p = h\rho g$ gives the TOTAL pressure (forgetting atmospheric)
Why it happens
Misreading the formula's meaning.
How to avoid it
Spec 5.7 formula gives the EXCESS pressure due to the FLUID COLUMN above. Total pressure on a submerged object = $p_\text{atm} + h\rho g$ . Read the question carefully.
✕Saying pressure 'pushes downwards' in a static fluid
Why it happens
Everyday image of weight pushing on the floor.
How to avoid it
Spec 5.6: pressure in a fluid at rest acts EQUALLY IN ALL DIRECTIONS at any point. The horizontal sides of a tank experience pressure too — that's why dams are built thicker at the base.
✕Confusing the volume of a regular solid with a circle/area formula
4PH1/1P Examiner Reports 2022-2024
Why it happens
Memorising formulas without practising.
How to avoid it
Cuboid: $V = lwh$ . Cylinder: $V = \pi r^2 h$ . Sphere: $V = \tfrac{4}{3}\pi r^3$ . For irregular shapes, use water displacement (spec 5.4).

Practice questions

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Density and Pressure — frequently asked questions

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Density and Pressure

Short Study Notes in the form of Slides

Short Study Notes — Density and Pressure

Detailed Notes

What you’ll learn

How it’s examined

1Density of a block (3 marks, Paper 1)

2Required practical: density of an irregular solid (6 marks, Paper 1)

3Pressure under feet (3 marks, Paper 1)

4Pressure at depth in a lake (4 marks, Paper 1)

5Pressure in a static fluid acts in all directions (3 marks, Paper 1)

Density (spec 5.3)

Pressure (spec 5.5)

Pressure difference in a fluid (spec 5.7)

Density (spec 5.3)

Pressure (spec 5.5)

Pressure direction in a static fluid (spec 5.6)

Pressure depth relationship (spec 5.7)

✕Mixing g/cm³ and kg/m³ in a single calculation

✕Quoting an area in cm² in the pressure formula

✕Saying p=hρgp = h\rho gp=hρg gives the TOTAL pressure (forgetting atmospheric)

✕Saying pressure 'pushes downwards' in a static fluid

✕Confusing the volume of a regular solid with a circle/area formula

Practice questions

Past paper style quiz

Assess your understanding

Video lesson

Density and Pressure — frequently asked questions

✕Saying $p = h\rho g$ gives the TOTAL pressure (forgetting atmospheric)