Density

Density is one of the most fundamental properties of matter. It tells you how much mass is packed into a given volume and is essential for understanding why some objects float while others sink, why materials are chosen for specific engineering purposes, and how to identify unknown substances.

Defining Density

Density is defined as mass per unit volume:

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

where:

ρ (Greek letter rho) is density, measured in kg m⁻³ (SI unit) or g cm⁻³
m is mass in kg (or g)
V is volume in m³ (or cm³)

To convert between the two common unit systems: 1000 kg m⁻³ = 1 g cm⁻³. Water has a density of approximately 1000 kg m⁻³ or 1.0 g cm⁻³.

Unit conversion trap: Many students lose marks by forgetting to convert cm³ to m³. Remember: 1 cm = 0.01 m, so 1 cm³ = (0.01)³ m³ = 1 × 10⁻⁶ m³. This means 1 m³ = 10⁶ cm³, not 100 cm³.

Reference Table: Densities of Common Materials

Material	Density / kg m⁻³	Density / g cm⁻³	State at room temp
Air (at sea level)	1.2	0.0012	Gas
Helium	0.164	0.000164	Gas
Cork	120	0.12	Solid
Softwood (pine)	500	0.50	Solid
Ice	917	0.917	Solid
Water	1000	1.00	Liquid
Seawater	1025	1.025	Liquid
Mercury	13 600	13.6	Liquid
Aluminium	2700	2.70	Solid
Glass	2500	2.50	Solid
Titanium	4500	4.50	Solid
Iron/Steel	7800	7.80	Solid
Copper	8900	8.90	Solid
Lead	11 300	11.3	Solid
Gold	19 300	19.3	Solid
Osmium (densest element)	22 590	22.6	Solid

These values explain material selection in engineering. Aircraft use aluminium alloys (low density, reasonable strength) rather than steel. Electrical wiring uses copper (excellent conductor) despite its relatively high density because conductivity matters more than weight in that application.

Measuring Density of Regular Solids

For objects with regular geometric shapes (cubes, cuboids, cylinders, spheres), you can calculate volume directly from measured dimensions.

Method:

Measure the mass using a digital balance (record in kg or g).
Measure the relevant dimensions using a ruler, vernier callipers, or micrometer.
Calculate volume using the appropriate formula:
- Cuboid: V = l × w × h
- Cylinder: V = πr²h
- Sphere: V = (4/3)πr³
Calculate density: ρ = m/V.

Worked example 1: A metal cylinder has mass 0.350 kg, diameter 2.40 cm, and height 5.00 cm.

Radius = 2.40/2 = 1.20 cm = 0.0120 m
Volume = π × (0.0120)² × 0.0500 = π × 1.44 × 10⁻⁴ × 0.0500 = 2.26 × 10⁻⁵ m³
Density = 0.350 / 2.26 × 10⁻⁵ = 15 500 kg m⁻³

This is close to the density of tungsten (19 300 kg m⁻³) or lead (11 300 kg m⁻³), suggesting it could be a lead alloy.

Worked example 2: A solid sphere has a mass of 4.50 kg and a diameter of 12.0 cm. Identify the material.

Radius = 0.060 m
Volume = (4/3)π(0.060)³ = (4/3)π × 2.16 × 10⁻⁴ = 9.047 × 10⁻⁴ m³
Density = 4.50 / 9.047 × 10⁻⁴ = 4973 kg m⁻³ ≈ 5000 kg m⁻³

Checking the table: this sits between titanium (4500) and iron (7800). It could be a titanium alloy or a composite material.

Measuring Density of Irregular Solids

When the shape is irregular and you cannot calculate volume from dimensions, you use displacement.

Method:

Measure the mass of the object on a balance.
Fill a measuring cylinder with water and record the initial volume V₁.
Carefully lower the object into the water (ensuring it is fully submerged and no air bubbles are trapped).
Record the new volume V₂.
Volume of object = V₂ − V₁.
Calculate density: ρ = m / (V₂ − V₁).

For larger objects, a eureka can (displacement can) can be used. Water overflows into a measuring cylinder through a spout when the object is submerged, and the collected water volume equals the object’s volume.

Key source of error: Surface tension can cause water to cling to the spout, so you should wait until dripping stops completely before reading the volume.

Uncertainty Analysis in Density Measurements

Understanding uncertainty is essential for practical work and exam questions.

Worked example 3 (with uncertainty): A student measures a cylinder: mass = 45.2 ± 0.1 g, diameter = 1.20 ± 0.02 cm, height = 8.50 ± 0.05 cm.

Step 1 — Calculate the density:

A = π(0.60)² = 1.131 cm²; V = 1.131 × 8.50 = 9.613 cm³
ρ = 45.2 / 9.613 = 4.70 g cm⁻³

Step 2 — Percentage uncertainties:

Mass: (0.1/45.2) × 100 = 0.22%
Height: (0.05/8.50) × 100 = 0.59%
Diameter: (0.02/1.20) × 100 = 1.67% → but area uses d², so % uncertainty in area = 2 × 1.67% = 3.33%

Step 3 — Total % uncertainty in density = 0.22% + 0.59% + 3.33% = 4.14%

Step 4 — Absolute uncertainty = 4.14% × 4.70 = ±0.19 g cm⁻³

Final answer: ρ = 4.70 ± 0.19 g cm⁻³ (or 4700 ± 190 kg m⁻³)

Exam tip: The diameter has the largest percentage uncertainty because it is the smallest measurement AND it is squared in the area calculation. This is a very common exam question.

Measuring Density of Liquids

To find the density of a liquid:

Place an empty measuring cylinder on a balance and record its mass m₁.
Pour a known volume V of liquid into the measuring cylinder.
Record the new mass m₂.
Mass of liquid = m₂ − m₁.
Density = (m₂ − m₁) / V.

Read the meniscus at eye level to avoid parallax error. For water and most liquids, read from the bottom of the meniscus. For mercury, read from the top.

Floating and Sinking

An object floats if its average density is less than the density of the fluid it is placed in. An object sinks if its average density is greater than the fluid density.

This explains why:

A steel ship floats — the hull encloses a large volume of air, making the ship’s average density much less than water.
Ice floats on water — ice has a density of about 917 kg m⁻³, which is less than water’s 1000 kg m⁻³.
A helium balloon rises — helium (0.164 kg m⁻³) is less dense than air (1.225 kg m⁻³).

When an object floats, it displaces a weight of fluid equal to its own weight. This is a direct consequence of Archimedes’ principle, which you will study in the next lesson.

Worked example 4: An iceberg of volume 500 m³ and density 917 kg m⁻³ floats in seawater (ρ = 1025 kg m⁻³). What volume is above the surface?

Fraction submerged = ρ_ice / ρ_seawater = 917 / 1025 = 0.8946
Volume submerged = 0.8946 × 500 = 447.3 m³
Volume above surface = 500 − 447.3 = 52.7 m³ (about 10.5%)

Density and the Particle Model

At the microscopic level, density depends on:

The mass of individual atoms or molecules — heavier atoms mean higher density.
How closely packed the particles are — tightly packed structures (like metals) have higher density.
The arrangement of particles — diamond (3500 kg m⁻³) and graphite (2200 kg m⁻³) are both pure carbon, but their different crystal structures give different densities.

In solids, particles are closely packed in fixed positions, giving high density. In gases, particles are widely spaced, giving very low density. Liquids fall in between, with densities typically similar to (but slightly less than) the corresponding solid — water being a notable exception where the solid (ice) is less dense than the liquid.

Alloy Density Calculation

When two materials are mixed to form an alloy, the resulting density can be estimated using volume fractions:

$\rho_{\text{alloy}} = f_1 \rho_1 + f_2 \rho_2$

where f₁ and f₂ are the volume fractions of each component.

Worked example 5: An alloy is 70% aluminium (ρ = 2700 kg m⁻³) and 30% copper (ρ = 8900 kg m⁻³) by volume.

ρ_alloy = 0.70 × 2700 + 0.30 × 8900 = 1890 + 2670 = 4560 kg m⁻³

Common Exam Mistakes

Forgetting volume unit conversion: 1 cm³ = 1 × 10⁻⁶ m³, not 1 × 10⁻² m³.
Using diameter instead of radius when calculating volumes of cylinders or spheres.
Confusing mass and weight in density calculations — density uses mass (kg), not weight (N).
Ignoring air bubbles when using displacement — trapped air reduces the apparent volume of the object, giving a density that is too high.
Assuming all objects of a material float or sink — it is the average density of the whole object (including air spaces) that matters, not just the density of the material itself.

A-Level Deep Dive: Density

Spec mapping

Edexcel 9PH0 Topic 4 — Materials, sub-topic on density requires students to be able to define density, recall and use the equation $\rho = m/V$ , and apply density to problems involving regular and irregular solids, liquids, and floating bodies (refer to the official Pearson Edexcel specification document for exact wording). Although density appears here as the gateway concept of Topic 4, it is genuinely synoptic: it is examined in 9PH0 Paper 1 (sections on Mechanics and Materials), Paper 2 (Thermodynamics through ideal-gas mass per molecule, and Astrophysics through stellar density), and the Practical Skills paper (CP2 Young modulus uses cross-sectional area derived from a measured diameter, exactly the kind of measurement chain assessed here). The Edexcel formula booklet provides $p = h \rho g$ but does not state $\rho = m/V$ — it must be memorised, along with the SI unit kg m⁻³.

Worked example with full mark scheme

Question (7 marks):

A cylindrical brass rod has a measured length of $L = 8.50 \pm 0.05$ cm and a measured diameter of $d = 1.20 \pm 0.02$ cm. Its mass is $m = 76.5 \pm 0.1$ g.

(a) Calculate the density of the rod, giving your answer in kg m⁻³. (3)

(b) Determine the absolute uncertainty in the density and quote your final answer to an appropriate number of significant figures. (4)

Solution with mark scheme:

(a) Step 1 — convert units and find the volume.

Convert to SI: $L = 0.0850$ m, $d = 0.0120$ m, so radius $r = 0.00600$ m. Mass $m = 0.0765$ kg.

Volume of a cylinder: $V = \pi r^2 L = \pi (0.00600)^2 (0.0850) = 9.613 \times 10^{-6}$ m³.

M1 — correct conversion of cm to m and use of radius (not diameter) in $\pi r^2 L$ . The most common slip is to use $d^2$ in place of $r^2$ , quadrupling the volume.

Step 2 — apply $\rho = m/V$ .

$\rho = \dfrac{0.0765}{9.613 \times 10^{-6}} = 7958\ \text{kg m}^{-3}$

M1 — substitution into $\rho = m/V$ with all quantities in SI units.

A1 — answer in the range 7950–7960 kg m⁻³, consistent with brass (typical brass density ≈ 8500 kg m⁻³, so the rod is at the low end of the brass range, possibly an unusual alloy).

(b) Step 1 — combine percentage uncertainties.

$\rho = m / (\pi r^2 L)$ , so the fractional uncertainties combine as:

$\dfrac{\Delta \rho}{\rho} = \dfrac{\Delta m}{m} + 2 \cdot \dfrac{\Delta d}{d} + \dfrac{\Delta L}{L}$

Note the factor of 2 on $\Delta d/d$ because diameter is squared in the volume formula (using diameter percentage equals using radius percentage, since the factor of 2 in $r = d/2$ cancels in fractional form).

M1 — recognising that $\Delta d/d$ is doubled.

Step 2 — substitute.

$\Delta m / m = 0.1/76.5 = 0.131\%$
$2 \Delta d / d = 2 \times (0.02/1.20) = 2 \times 1.67\% = 3.33\%$
$\Delta L / L = 0.05/8.50 = 0.588\%$

Total: $0.131 + 3.33 + 0.588 = 4.05\%$ .

M1 — correct addition of fractional uncertainties.

Step 3 — convert to absolute uncertainty.

$\Delta \rho = 4.05\% \times 7958 = 322$ kg m⁻³, rounded to 1 significant figure for the uncertainty: $\pm 300$ kg m⁻³.

A1 — absolute uncertainty within ±20 of 320 kg m⁻³ (or quoted to 1 sf as 300).

Step 4 — quote sensibly.

The uncertainty has two significant figures of precision at most. The mean must be quoted to a precision compatible with the uncertainty: $\rho = (8.0 \pm 0.3) \times 10^3$ kg m⁻³.

A1 — final value and uncertainty quoted with consistent precision; the standard-form presentation is rewarded as showing examiner-aware significant-figure discipline.

Total: 7 marks (M3 A4, split as shown).

Specimen question modelled on the Edexcel 9PH0 Paper 1 format

Question (6 marks): A rectangular metal block has dimensions $5.00 \times 4.00 \times 2.00$ cm and a mass of 312 g.

(a) Calculate its density in kg m⁻³, and identify the most likely metal from the table below. (3)

Metal	Density / kg m⁻³
Aluminium	2700
Iron	7800
Copper	8900
Silver	10 500
Lead	11 300

(b) The block is then drilled to remove a cylindrical core of diameter 1.00 cm right through its 5.00 cm length. Calculate the new density of the drilled block. (3)

Mark scheme decomposition by AO:

(a)

M1 (AO2.1) — converting dimensions to SI and computing $V = 5.00 \times 4.00 \times 2.00 = 40.0$ cm³ $= 4.00 \times 10^{-5}$ m³, with mass $m = 0.312$ kg.
A1 (AO1.1b) — $\rho = 0.312 / (4.00 \times 10^{-5}) = 7800$ kg m⁻³.
B1 (AO3.2a) — identifying the metal as iron based on the density match.

(b)

M1 (AO2.1) — recognising that mass and volume both decrease: new volume $= 4.00 \times 10^{-5} - \pi (0.005)^2 (0.05) = 4.00 \times 10^{-5} - 3.93 \times 10^{-6} = 3.61 \times 10^{-5}$ m³.
M1 (AO2.1) — new mass $= 0.312 - (7800 \times 3.93 \times 10^{-6}) = 0.312 - 0.0306 = 0.281$ kg, assuming the removed material has the same density as the rest.
A1 (AO1.1b) — new density $= 0.281 / 3.61 \times 10^{-5} = 7800$ kg m⁻³.

The "trick" of part (b) is realising that drilling a uniform material does not change its density: removing equal proportions of mass and volume leaves the ratio unchanged. Many candidates compute only the new volume, dividing the original mass by it, and get a wrong (higher) density.

Total: 6 marks split AO1 = 2, AO2 = 3, AO3 = 1. This is a synoptic-style application question that tests measurement, manipulation, and conceptual reasoning together.

Synoptic links

Connects to:

Topic 2 — Mechanics (forces and equilibrium): density underpins the calculation of weight $W = mg = \rho V g$ for any uniform body. Free-body diagrams of submerged or floating objects use $\rho V g$ as the weight term and $\rho_{\text{fluid}} V_{\text{disp}} g$ as the upthrust — the next lesson formalises this via Archimedes.
Topic 5 — Waves and the particle nature of light (refraction): the optical density of a medium correlates with mass density through the Lorentz–Lorenz relation; denser materials generally have higher refractive indices, which is why glass ( $\rho \approx 2500$ kg m⁻³, $n \approx 1.5$ ) refracts more strongly than water ( $\rho = 1000$ kg m⁻³, $n = 1.33$ ).
Topic 9 — Thermodynamics (kinetic theory): for an ideal gas, $\rho = pM/(RT)$ where $M$ is molar mass — an expression derived from $pV = nRT$ by substituting $n = m/M$ and rearranging $\rho = m/V$ . This is exactly how astronomers compute stellar atmospheric density from spectroscopic temperature and pressure.
Topic 12 — Gravitational fields: the average density of a planet $\rho = M/V$ governs its surface gravitational field strength $g = GM/R^2 = (4/3)\pi G \rho R$ . Earth's average density of 5510 kg m⁻³ — much higher than surface rock — is the principal evidence for a dense iron core.
Practical Skills (CP2 Young modulus): measuring the density of a wire is conceptually identical to measuring its cross-sectional area, the bottleneck step in Young modulus determination. The percentage-uncertainty arithmetic — that $\Delta d/d$ is doubled because $d^2$ appears — transfers directly.

Mark-scheme literacy

Density questions on 9PH0 split AO marks as follows:

AO	Typical share	Earned by
AO1 (knowledge / procedure)	50–60%	Recalling $\rho = m/V$ , computing volumes of standard shapes, converting units, applying the equation in routine substitution
AO2 (application / reasoning)	25–35%	Choosing the appropriate volume formula, combining percentage uncertainties, identifying material from density, recognising when density is unchanged after removal of uniform material
AO3 (problem-solving / experimental)	10–25%	Designing a displacement protocol, evaluating dominant uncertainty sources, suggesting improvements (vernier vs ruler, mean of repeated readings)

Examiner-rewarded phrasing: "converting cm³ to m³ using $1\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^3$ "; "the percentage uncertainty in $d^2$ is twice the percentage uncertainty in $d$ "; "the mean and uncertainty are quoted to consistent precision". Phrases that lose marks: "density is mass times volume" (definition error — costs the entire B1); leaving an answer in g cm⁻³ when the question demands kg m⁻³; quoting an uncertainty to four significant figures (e.g. ±322.4 kg m⁻³) when the data justifies one.

A specific Edexcel pattern to watch: when a question says "show that the density is approximately X", the candidate is expected to compute a value, state explicitly that it agrees with X, and only then continue to subsequent parts. Skipping the explicit comparison forfeits the B1 even if the numerical work is flawless.

Grade-band model answers

3-mark question

Question: A glass marble has a mass of 5.20 g. When dropped into a measuring cylinder containing 25.0 cm³ of water, the water level rises to 27.1 cm³. Calculate the density of the glass in kg m⁻³.

Grade C response (~150 words):

Volume of marble = 27.1 − 25.0 = 2.1 cm³.

Density = 5.20 / 2.1 = 2.476 g cm⁻³.

Converting: 2.476 × 1000 = 2476 kg m⁻³.

Examiner commentary: Full marks (3/3). The displacement subtraction is correctly done, the density formula is applied, and the unit conversion is correct. This is a clean Grade C answer — every step is shown, but the candidate quotes the volume to 2 sf only because the data justifies no more (a subtlety worth noting: 27.1 − 25.0 = 2.1 is not 2.10, because the resolution of the measurement caps the precision). The final density at 2476 kg m⁻³ is in the right region for glass (typical 2400–2600 kg m⁻³), so the candidate could optionally note this match for an extra layer of confidence — examiners do not award marks for the verification but do mark down internally inconsistent answers.

Grade A response (~210 words):*

The volume of the marble equals the volume of water displaced:

$V = V_2 - V_1 = 27.1 - 25.0 = 2.1\ \text{cm}^3 = 2.1 \times 10^{-6}\ \text{m}^3$

Converting the mass to SI: $m = 5.20$ g $= 5.20 \times 10^{-3}$ kg.

Applying the density definition:

$\rho = \dfrac{m}{V} = \dfrac{5.20 \times 10^{-3}}{2.1 \times 10^{-6}} = 2476\ \text{kg m}^{-3} \approx 2.5 \times 10^3\ \text{kg m}^{-3}$

Reported to 2 sf, $\rho = 2.5 \times 10^3$ kg m⁻³, consistent with the expected density of soda-lime glass.

Examiner commentary: Full marks (3/3). Beyond procedural correctness, the candidate does three things examiners reward: (i) converts everything to SI units before substituting (avoiding the error-prone step of mixing units mid-calculation), (ii) quotes the answer to a number of significant figures consistent with the data (the volume is only known to 2 sf, so the density cannot be more precise), and (iii) closes with a sanity check — naming "soda-lime glass" demonstrates the candidate is interpreting the number, not just generating it. This habit transfers to harder questions where verifying an answer against expected scale catches arithmetic slips.

6-mark question

Question: A student investigates the density of an irregularly shaped pebble using displacement. The pebble has mass $m = 75.4 \pm 0.1$ g. When submerged in a measuring cylinder, the water level rises from $V_1 = 50.0 \pm 0.5$ cm³ to $V_2 = 78.0 \pm 0.5$ cm³. (a) Calculate the density of the pebble in kg m⁻³ with its absolute uncertainty. (b) Suggest one improvement to the experimental method that would significantly reduce the dominant uncertainty. (6)

Grade B response (~280 words):

(a) Volume of pebble = 78.0 − 50.0 = 28.0 cm³ = 2.80 × 10⁻⁵ m³. Mass = 0.0754 kg. Density = 0.0754 / 2.80 × 10⁻⁵ = 2693 kg m⁻³.

Uncertainty in volume: when subtracting, absolute uncertainties add: $\Delta V = 0.5 + 0.5 = 1.0$ cm³, so $\Delta V/V = 1.0/28.0 = 3.57\%$ . Uncertainty in mass: 0.1/75.4 = 0.13%. Total: 3.70%. Absolute = 3.70% × 2693 = 100 kg m⁻³.

So $\rho = 2700 \pm 100$ kg m⁻³.

(b) Use a more precise measuring cylinder with smaller graduations.

Examiner commentary: Part (a) is correct (4/4) — the candidate correctly adds absolute uncertainties on subtraction (a step many forget) and combines fractional uncertainties on division. The final value and its uncertainty are quoted to consistent precision. Part (b) earns 1/2 — "more precise measuring cylinder" is the right idea but generic. To gain the second mark, the candidate would need to specify why: the volume uncertainty dominates (3.57% versus 0.13% for mass), so reducing it has the largest impact, and a typical 100 cm³ cylinder reads to ±0.5 cm³, while a 50 cm³ cylinder reads to ±0.25 cm³ — halving the dominant error term. Total: 5/6.

Grade A response (~340 words):*

(a) Volume of pebble (by displacement):

$V = V_2 - V_1 = 78.0 - 50.0 = 28.0\ \text{cm}^3 = 2.80 \times 10^{-5}\ \text{m}^3$

When subtracting two measurements, absolute uncertainties add:

$\Delta V = \Delta V_1 + \Delta V_2 = 0.5 + 0.5 = 1.0\ \text{cm}^3$

Fractional uncertainty in volume: $\Delta V/V = 1.0/28.0 = 3.57\%$ .

Fractional uncertainty in mass: $\Delta m/m = 0.1/75.4 = 0.13\%$ .

Density: $\rho = m/V = 0.0754 / 2.80 \times 10^{-5} = 2693$ kg m⁻³.

Total fractional uncertainty (sum, since $\rho = m/V$ involves division): $3.57\% + 0.13\% = 3.70\%$ .

Absolute uncertainty: $3.70\% \times 2693 = 100$ kg m⁻³.

Final answer: $\rho = (2.7 \pm 0.1) \times 10^3$ kg m⁻³.

(b) The volume uncertainty (3.57%) dominates the density uncertainty by a factor of ~30 over the mass uncertainty. To reduce it significantly, replace the 100 cm³ measuring cylinder (resolution ±0.5 cm³) with a 50 cm³ cylinder reading to ±0.25 cm³, or — better — use a Eureka can to displace the water into a 25 cm³ cylinder graduated to ±0.1 cm³, dropping the absolute volume uncertainty from 1.0 to ~0.2 cm³. This reduces the dominant fractional uncertainty from 3.57% to ~0.7%, bringing the total uncertainty in density from 3.70% to ~0.9%.

Examiner commentary: Full marks (6/6). The candidate identifies the dominant error term quantitatively (3.57% vs 0.13%), then proposes a method change and estimates its impact (3.57% → ~0.7%). This is exactly the AO3 evaluative reasoning Edexcel rewards — not just naming an improvement, but justifying it numerically. The standard-form quoting of the final answer with consistent significant figures shows mature presentation discipline.

9-mark question

Question: A solid composite cube has side length $a = 5.00$ cm and total mass $m = 437$ g. The cube is made of two materials, A and B, fused along a horizontal plane: layer A occupies the lower 60% of the cube's height, with density $\rho_A = 8000$ kg m⁻³, and layer B occupies the upper 40%, with unknown density $\rho_B$ .

(a) Calculate $\rho_B$ . (4)

(b) The composite cube is placed in water (ρ = 1000 kg m⁻³). State, with justification, whether the cube floats or sinks. (2)

(c) If the cube floats, what fraction of its height projects above the water? If it sinks, what is its weight in water (i.e. apparent weight)? Take $g = 9.81$ m s⁻². (3)

Grade A response (~370 words):*

(a) Volume of layer A: $V_A = a^2 \times 0.6a = 0.6 \times (0.05)^3 = 7.5 \times 10^{-5}$ m³.

Volume of layer B: $V_B = 0.4 \times (0.05)^3 = 5.0 \times 10^{-5}$ m³.

Mass of A: $m_A = \rho_A V_A = 8000 \times 7.5 \times 10^{-5} = 0.600$ kg.

Total mass: $m = 0.437$ kg. Mass of B: $m_B = 0.437 - 0.600 = -0.163$ kg.

A negative mass is unphysical — re-checking the question, total mass must exceed $m_A$ . Assuming the question intends $m = 0.737$ kg (so $m_B = 0.137$ kg), then:

$\rho_B = \dfrac{m_B}{V_B} = \dfrac{0.137}{5.0 \times 10^{-5}} = 2740\ \text{kg m}^{-3}$

(b) Average density: $\rho_{\text{avg}} = m/V_{\text{total}} = 0.737 / (1.25 \times 10^{-4}) = 5896$ kg m⁻³, much greater than water's 1000 kg m⁻³. The cube sinks.

True weight: $W = mg = 0.737 \times 9.81 = 7.23$ N.

Upthrust (weight of water displaced, since fully submerged): $U = \rho_{\text{water}} V g = 1000 \times 1.25 \times 10^{-4} \times 9.81 = 1.23$ N.

Apparent weight: $W' = 7.23 - 1.23 = 6.00$ N.

Examiner commentary: This question reveals a printing inconsistency in the stem (with $m = 437$ g, $\rho_B$ would be negative). The candidate handles this with the level of care examiners reward at A*: they identify the inconsistency explicitly, propose the most likely intended value, and continue with a stated assumption rather than producing nonsense. In the actual exam, raising your hand for a clarification is correct, but in past-paper practice, "stating an assumption clearly and proceeding" earns full follow-through marks. The physical reasoning in (b) and (c) is exemplary: average-density argument for floatation is exactly what mark schemes expect, and the Archimedes upthrust calculation in (c) is a perfect synoptic preview of the next lesson. Full marks where the question is unambiguous; flagged inconsistency in (a). Total: 8–9/9 depending on credit for the clarification.

A-Level-depth misconceptions

The errors that distinguish A from A* on density questions:

Density is not a function of size. A 1 cm³ block of iron has the same density as a 1 m³ block of iron — both are 7800 kg m⁻³. Yet candidates regularly write "the larger block has higher density" or, more commonly, conflate density with mass: "iron is heavier than aluminium" (correct only at equal volume).
Average density vs material density. A steel ship has steel density 7800 kg m⁻³ but average density much less than 1000 kg m⁻³ (otherwise it would sink). The float/sink criterion uses average density of the whole object, not the density of the material it is made from.
The $d^2$ trap in cross-section. When measuring a cylindrical wire's volume from $V = \pi r^2 L = \pi (d/2)^2 L = \pi d^2 L/4$ , the diameter is squared. The percentage uncertainty in $d^2$ is twice the percentage uncertainty in $d$ , not the same. This single error is the most common reason candidates get density-uncertainty calculations wrong.
Unit conversion: 1 cm³ ≠ 0.01 m³. Candidates know 1 cm = 0.01 m and incorrectly cube only the unit, not the prefix: $1\ \text{cm}^3 = (10^{-2}\ \text{m})^3 = 10^{-6}$ m³. Forgetting the cubing of $10^{-2}$ produces densities off by a factor of $10^4$ .
Density of a hollow object is not the density of its material. A hollow steel sphere has the density of steel only on a per-unit-of-actual-steel basis. Treating the sphere's enclosed volume (including the air cavity) as the volume in $\rho = m/V$ gives a much lower number — which is physically the relevant one for floatation.
Specific gravity vs density. "Specific gravity" (relative density) is the ratio of an object's density to water's, and is dimensionless. Candidates sometimes treat a quoted specific gravity of 2.5 as "2.5 kg m⁻³" — off by a factor of 1000 from the true 2500 kg m⁻³.
Temperature dependence is not zero. Density depends weakly on temperature for solids and liquids (volumes change with thermal expansion) and strongly for gases ( $\rho \propto 1/T$ at constant pressure). At A-Level, you are expected to know density is temperature-dependent for gases, even if numerical questions usually assume room temperature.

Common errors and mark-loss patterns on density measurement

Three patterns repeatedly cost candidates marks on Paper 1 density questions. They are all about presentation and unit discipline, not concept.

Mid-calculation unit drift. Candidates start in g and cm, switch to kg and m partway through, and lose track of where the conversion factor of $10^3$ or $10^6$ should appear. The cure: convert everything to SI on the first line, then never touch units again until the final answer.
Significant-figure inflation. Quoting $\rho = 2476.351$ kg m⁻³ when the inputs were measured to 3 sf is wrong; the answer cannot be more precise than the data. The final A1 in many density questions is awarded for the quoted value matching the precision of the inputs.
Volume formula confusion. Cylinder volume is $\pi r^2 L$ , not $\pi d^2 L$ or $2\pi r L$ . The latter is the surface area; the former misses the factor of 4. Examiners see all three errors regularly, and any of them invalidates the density value at the M1 stage, cascading errors of factor 4 or more through the rest of the answer.

Going further — university and academic signposting

Density and the closely related concept of mass distribution point directly toward several undergraduate trajectories:

Continuum mechanics (Year 2): the local density $\rho(\vec{r})$ becomes a field — a function of position — and the total mass is the volume integral $m = \int \rho\,dV$ . Conservation of mass in a flowing fluid is then the continuity equation $\partial \rho/\partial t + \nabla \cdot (\rho \vec{v}) = 0$ , foundational to fluid dynamics and aerodynamics.
Astrophysics (Year 2/3): stellar structure depends on the radial density profile $\rho(r)$ . The Lane–Emden equation, derived from hydrostatic equilibrium and a polytropic equation of state $p = K\rho^{(n+1)/n}$ , predicts the density profile of stars and is the origin of the Chandrasekhar limit on white-dwarf mass.
Solid-state physics (Year 2/3): packing fractions of crystal structures (FCC: 74%, BCC: 68%, simple cubic: 52%) directly determine the density of the bulk material from atomic mass and lattice parameter. The X-ray diffraction technique allows you to measure lattice constants to ~5 decimal places and predict density to comparable precision.
Cosmology (Year 3/4): the critical density of the universe, $\rho_c = 3H_0^2/(8\pi G) \approx 9 \times 10^{-27}$ kg m⁻³, determines whether the universe is open, flat, or closed. Measurements of the cosmic microwave background pin the actual density to within a few percent of $\rho_c$ .

Oxbridge interview prompt: "Estimate the average density of a neutron star, given that it has the mass of the Sun packed into a sphere of radius 10 km. How does that compare to atomic density, and what does the comparison tell you about the structure of the neutrons inside?"

Required practical reference

Density is the central measurement skill underpinning CP2 — determination of the Young modulus of a metal wire. To find the Young modulus $E = \sigma/\varepsilon = (FL)/(A \Delta L)$ , the cross-sectional area $A = \pi d^2/4$ is computed from a micrometer measurement of the wire's diameter at three or more points along its length. The diameter is the dominant error term — typically $d \sim 0.4$ mm with a micrometer resolution of ±0.01 mm, giving fractional uncertainty of ~2.5%, doubled to 5% in $d^2$ . This single measurement contributes more uncertainty than load, length, and extension combined, exactly as in the density-of-a-rod calculation worked above. CP2 also implicitly tests density: students who do not recognise that uniform drilling or stretching does not change density will mis-handle compound questions about volume conservation under elastic strain. CP4 — viscosity of a fluid by Stokes' Law uses density twice: the density $\rho_s$ of the falling sphere appears in its weight $W = \rho_s V g$ , and the density $\rho_f$ of the surrounding fluid appears in the upthrust $U = \rho_f V g$ , with terminal velocity governed by the difference $(\rho_s - \rho_f)$ . CP9 — investigating absolute zero through gas pressure–temperature relations makes the gas-density connection explicit: at constant volume, $p \propto \rho T/M$ , and extrapolation to zero pressure gives absolute zero. Each of these practicals re-uses the percentage-uncertainty arithmetic introduced here: master it now and the rest of the practical paper becomes routine.

Edexcel A-Level alignment footer

This content is aligned with the Pearson Edexcel GCE A Level Physics (9PH0) specification, Topic 4 — Materials, sub-topic on density. For the most accurate and up-to-date information, please refer to the official Pearson Edexcel specification document.

Visual summary

graph TD
    A["Object whose density<br/>is to be determined"] --> B{"Shape?"}
    B -->|"Regular<br/>(cube, cylinder, sphere)"| C["Measure dimensions<br/>with vernier or micrometer<br/>V from formula"]
    B -->|"Irregular solid"| D["Displacement method:<br/>V = V₂ − V₁<br/>(Eureka can if large)"]
    B -->|"Liquid"| E["Mass of liquid =<br/>m₂ − m₁ in cylinder<br/>V from cylinder reading"]
    C --> F["Measure mass<br/>on digital balance"]
    D --> F
    E --> F
    F --> G["Apply<br/>ρ = m/V<br/>in SI units"]
    G --> H["Combine fractional<br/>uncertainties:<br/>doubling Δd/d<br/>where d is squared"]
    H --> I{"Quote with<br/>matching precision"}
    I --> J["Final ρ in kg m⁻³<br/>± absolute uncertainty"]

    style G fill:#27ae60,color:#fff
    style J fill:#3498db,color:#fff