What is density and how is it calculated in the context of Edexcel International A Levels Physics?

Density is a measure of how much mass is contained in a given volume. It is calculated using the formula: Density (ρ) = Mass (m) / Volume (V). In the Edexcel International A Levels Physics curriculum, understanding density is crucial for solving problems related to Mechanics and Materials, as it helps in determining the buoyancy and stability of objects in fluids.

How does upthrust affect an object submerged in a fluid according to Edexcel International A Levels Physics?

Upthrust, also known as buoyant force, is the upward force exerted by a fluid on an object submerged in it. According to Archimedes' principle, the upthrust is equal to the weight of the fluid displaced by the object. In Edexcel International A Levels Physics, understanding upthrust is essential for analyzing the floating and sinking behavior of objects, which is a key concept in the study of Mechanics and Materials.

What factors influence viscous drag on an object moving through a fluid in Edexcel International A Levels Physics?

Viscous drag is the resistance experienced by an object moving through a fluid. It is influenced by several factors, including the object's speed, the fluid's viscosity, and the object's shape and surface area. In Edexcel International A Levels Physics, students learn that higher viscosity and speed increase viscous drag, while streamlined shapes can reduce it, which is important for understanding fluid dynamics in Mechanics and Materials.

How do you differentiate between density and upthrust in the context of Edexcel International A Levels Physics?

Density and upthrust are related but distinct concepts. Density refers to the mass per unit volume of a substance, while upthrust is the buoyant force exerted by a fluid on an object submerged in it. In Edexcel International A Levels Physics, understanding the difference is crucial for solving problems involving buoyancy, as density determines whether an object will float or sink, while upthrust determines the net force acting on the object in a fluid.

Why is understanding viscous drag important in the study of Mechanics and Materials for Edexcel International A Levels Physics?

Understanding viscous drag is important because it affects the motion of objects through fluids, which is a key topic in Mechanics and Materials. In Edexcel International A Levels Physics, students learn how viscous drag influences the terminal velocity of falling objects, the efficiency of vehicles, and the behavior of fluids in various applications. Mastery of this concept helps in predicting and optimizing the performance of systems involving fluid flow.

Why is "Using the density of the OBJECT instead of the density of the FLUID in $U = \rho V g$" a common mistake in Density, Upthrust and Viscous Drag?

Why it happens: Students see one object and one density value and use whatever they are given, without thinking about which density Archimedes' principle requires. How to avoid it: Upthrust equals the weight of **fluid displaced**, so always use the **density of the fluid** with the submerged volume. Write U = \rho_f V g with the subscript f as a reminder. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Forgetting upthrust at terminal velocity (writing weight = viscous drag instead of weight = upthrust + drag)" a common mistake in Density, Upthrust and Viscous Drag?

Why it happens: Air-resistance problems from earlier study only balance weight against drag, so students carry that habit into denser fluids where upthrust is significant. How to avoid it: In a liquid the upthrust is NOT negligible. Always use **weight = upthrust + viscous drag**, which gives the density DIFFERENCE (\rho_s - \rho_f) in the terminal-velocity formula. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Putting the diameter into Stokes' law instead of the radius" a common mistake in Density, Upthrust and Viscous Drag?

Why it happens: Ball-bearing sizes are quoted as diameters and measured with a micrometer as a diameter, so the value to hand is the diameter. How to avoid it: Stokes' law uses the **radius**: halve a measured diameter first. Because the terminal-velocity formula contains r^2, this error changes the result by a factor of four. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Leaving the radius in mm or the volume in cm³ when substituting into SI equations" a common mistake in Density, Upthrust and Viscous Drag?

Why it happens: Data are often given in convenient lab units (mm, cm³, g cm⁻³) and students substitute them straight in. How to avoid it: Convert everything to **SI base units first**: mm → m (× 10^-3), cm³ → m³ (× 10^-6), g cm⁻³ → kg m⁻³ (× 1000). Then the answer comes out in N, m s⁻¹ or Pa s as required. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Applying Stokes' law to large/fast objects or turbulent flow" a common mistake in Density, Upthrust and Viscous Drag?

Why it happens: Students treat F = 6 r v as a universal drag formula rather than a special case. How to avoid it: Stokes' law holds **only for a small, smooth sphere in laminar flow**. For large or fast objects the flow is turbulent and the formula does not apply — state this assumption in evaluation questions. Source: Edexcel IAL Physics Unit 1 examiner reports

Why is "Treating viscosity as constant and ignoring temperature in the falling-ball experiment" a common mistake in Density, Upthrust and Viscous Drag?

Why it happens: Viscosity is quoted as a single number in data books, so students assume it does not change during the experiment. How to avoid it: Viscosity of a liquid **decreases as temperature rises** (gases are the opposite), so the temperature must be **kept constant and recorded** (e.g. a water bath). Quote any viscosity value with the temperature it was measured at. Source: Edexcel IAL Physics Unit 1 examiner reports

Edexcel International A Levels Physics (XPH11-YPH11)

Density, Upthrust and Viscous Drag

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Short Notes - Density, Upthrust and Viscous Drag

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Detailed Study Notes

Detailed notes on Mechanics and Materials for Edexcel International A Levels Physics, covering key concepts, explanations, examples, and exam-focused revision points.

Density, Upthrust and Viscous Drag — Pearson Edexcel International A Level Physics (XPH11/YPH11) Study Notes (Unit 1)

Density $\rho = m/V$ , Archimedes' upthrust as the weight of fluid displaced, floating versus sinking by comparing densities, Stokes' law for viscous drag in laminar flow, and the terminal velocity of a sphere falling through a fluid — including Core Practical 2, the falling-ball method for measuring viscosity.

At a glance

Density is mass per unit volume: $\rho = m/V$ , in $\text{kg m}^{-3}$ . Remember $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ .
Upthrust = weight of fluid displaced (Archimedes' principle): $U = \rho_{f} V g$ , using the fluid's density and the submerged volume.
An object floats if its average density is less than the fluid's; it sinks if its average density is greater.
When floating, the upthrust on the submerged part equals the weight — the fraction submerged $= \rho_{object}/\rho_{fluid}$ .
Viscosity $\eta$ (units $\text{Pa s}$ ) measures a fluid's resistance to flow; honey is more viscous than water.
Stokes' law gives the viscous drag on a small smooth sphere in laminar flow: $F = 6\pi\eta r v$ (uses the radius, not diameter).
At terminal velocity the resultant force is zero: weight = upthrust + viscous drag, giving $v_{t} = \dfrac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$ .
Core Practical 2 finds viscosity by timing a ball at terminal velocity between two marks and rearranging Stokes' law — temperature must be controlled.

What you’ll learn

Mapped to the Pearson Edexcel International A Levels XPH11-YPH11 syllabus (2018-onwards).

Use the relationship $\rho = m/V$ to calculate density, including the average density of composite or hollow objects.
Recall Archimedes' principle and use $U = \rho_{f} V g$ to calculate upthrust as the weight of fluid displaced.
Predict and explain whether an object floats or sinks by comparing its (average) density with that of the fluid, and by comparing weight with upthrust.
Describe viscosity as a measure of resistance to flow and recall that the viscosity of a liquid decreases with temperature (gases increase).
Use Stokes' law $F = 6\pi\eta rv$ for the viscous drag on a small sphere in laminar flow, and recognise its conditions of validity.
Analyse the motion of a sphere falling through a fluid to terminal velocity using the force balance weight = upthrust + viscous drag.
Core Practical 2: determine the viscosity of a liquid using the falling-ball method, and evaluate the experiment.

Density — mass packed into a volume

The starting point: how much mass occupies a given volume.

Density is the mass per unit volume of a material:

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

where $\rho$ (rho) is density, $m$ is mass and $V$ is volume. The SI unit is the kilogram per cubic metre, $\text{kg m}^{-3}$ .

You will often meet density quoted in $\text{g cm}^{-3}$ . To convert, remember the volume factor is cubed:

$1\ \text{g cm}^{-3} = \frac{1 \times 10^{-3}\ \text{kg}}{1 \times 10^{-6}\ \text{m}^3} = 1000\ \text{kg m}^{-3}.$

So water ( $1.00\ \text{g cm}^{-3}$ ) has density $1000\ \text{kg m}^{-3}$ ; aluminium is $2700\ \text{kg m}^{-3}$ ; steel about $7800\ \text{kg m}^{-3}$ .

For a composite or hollow object (a ship, a submarine, a bottle of air), what matters is the average density — the total mass divided by the total volume. A steel ship is made of a material far denser than water, yet its average density (steel hull plus the air inside) is less than that of water, so it floats. Always ask: is it the density of the material, or the average density of the whole object, that the question needs?

$\rho = m/V$ , unit $\text{kg m}^{-3}$ .
$1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ (the cm→m factor is cubed).
Average density = total mass ÷ total volume — this decides floating/sinking.

See the full worked example for density, upthrust and viscous drag →

Upthrust — Archimedes' principle

Why a fluid pushes objects up: the weight of fluid displaced.

When an object is placed in a fluid (a liquid or a gas), the fluid pushes up on it with a force called upthrust (or the buoyancy force). Upthrust arises because pressure increases with depth: the fluid pressure on the bottom of the object is greater than the pressure on its top, giving a resultant upward force.

Archimedes' principle states this precisely:

The upthrust on a body immersed in a fluid is equal to the weight of the fluid displaced by the body.

The displaced fluid has volume equal to the submerged volume of the object, so

$U = \rho_{f} V g$

where $\rho_{f}$ is the density of the fluid (not the object), $V$ is the volume of fluid displaced, and $g$ is the gravitational field strength. The subscript "f" is your reminder always to use the fluid's density.

Fluid pressure is greater at the bottom of the object than at the top, so the upward force on the base exceeds the downward force on the top. The resultant is the upthrust, equal to the weight of fluid displaced ($U = \rho_{f} V g$).

The same idea works in air, but air is so much less dense than most solids that its upthrust is usually negligible — which is why we can normally ignore upthrust for objects falling through air, but never for objects in a liquid.

Upthrust comes from the pressure difference (top vs bottom).
Archimedes: upthrust = weight of fluid displaced.
$U = \rho_{f} V g$ — use the FLUID density and the submerged volume.

See the full worked example for density, upthrust and viscous drag →

Floating or sinking — compare the densities

An object floats if it is less dense than the fluid; the fraction submerged is a density ratio.

Whether an object floats or sinks comes down to a single comparison.

If the object is fully submerged, the upthrust is the weight of an equal volume of fluid. Compare this with the object's weight:
- Average density of object < fluid density → upthrust > weight → resultant force up → it rises and floats.
- Average density of object > fluid density → weight > maximum upthrust → resultant force down → it sinks.
- Equal densities → it stays where it is put (neutral buoyancy).

When an object floats, it rises until just enough of it is submerged that the upthrust exactly equals the weight. Equating these for a floating object,

$\rho_{object} V g = \rho_{fluid} V_{sub}\, g \quad\Rightarrow\quad \frac{V_{sub}}{V} = \frac{\rho_{object}}{\rho_{fluid}},$

so the fraction submerged equals the ratio of densities. An iceberg ( $917\ \text{kg m}^{-3}$ ) in sea water ( $1025\ \text{kg m}^{-3}$ ) floats with $917/1025 = 0.895$ , i.e. about 89.5% below the surface — the famous "tip of the iceberg".

Left: a less-dense object floats, with the upthrust on the submerged part balancing its weight ($U = W$). Right: a denser object's weight exceeds the maximum upthrust, so there is a resultant downward force and it sinks.

Float if average density < fluid density; sink if average density > fluid density.
Floating: upthrust on submerged part = weight ( $U = W$ ).
Fraction submerged $= \rho_{object}/\rho_{fluid}$ (g and V cancel).

See the full worked example for density, upthrust and viscous drag →

Viscosity and Stokes' law — drag in a fluid

How fluids resist motion, and the drag on a small sphere in laminar flow.

When an object moves through a fluid, the fluid exerts a backward viscous drag (frictional) force on it. How large this drag is depends on the fluid's viscosity, $\eta$ (Greek "eta").

Viscosity measures a fluid's resistance to flow — how readily one layer of fluid slides over the next. Honey has a high viscosity and flows slowly; water has a low viscosity. The SI unit of viscosity is the pascal second, $\text{Pa s}$ (equivalently $\text{kg m}^{-1}\text{s}^{-1}$ ).

Temperature matters. For a liquid, viscosity decreases as temperature rises (the molecules gain kinetic energy and slide past each other more easily — warm oil runs more freely). For a gas, viscosity increases with temperature. This is why temperature must be controlled in any viscosity experiment.

For a small, smooth sphere moving slowly through a fluid so that the flow is laminar (streamline), the viscous drag is given by Stokes' law:

$F = 6\pi\eta r v$

where $r$ is the radius of the sphere (not its diameter) and $v$ is its speed relative to the fluid. The drag is directly proportional to speed — so as a falling object speeds up, the drag on it grows.

Stokes' law is valid only when:

the flow is laminar, not turbulent;
the sphere is small and smooth, moving slowly;
the fluid is effectively unbounded (no significant wall effects from a narrow container).

For large or fast objects the flow becomes turbulent, eddies form, and Stokes' law no longer applies.

Viscosity $\eta$ = resistance to flow; unit $\text{Pa s}$ .
Liquids: viscosity falls as temperature rises. Gases: it rises.
Stokes' law $F = 6\pi\eta r v$ — small smooth sphere, laminar flow, radius (not diameter).
Drag $\propto$ speed, so it grows as the object accelerates.

See the full worked example for density, upthrust and viscous drag →

Terminal velocity and Core Practical 2

Weight = upthrust + viscous drag; using a falling ball to measure viscosity.

Drop a small sphere into a tall column of liquid and three forces act on it: its weight $W$ (down), the upthrust $U$ (up), and the viscous drag $F$ (up, opposing the motion).

Just after release the sphere moves slowly, so the drag is small. The weight exceeds the upthrust plus drag, giving a resultant downward force, so the sphere accelerates. As it speeds up, the drag increases (because $F = 6\pi\eta r v$ grows with $v$ ), the resultant force shrinks, and the acceleration falls. Eventually the sphere reaches a speed where

$\underbrace{W}_{\text{weight}} = \underbrace{U}_{\text{upthrust}} + \underbrace{F}_{\text{viscous drag}}$

The resultant force is now zero, so the acceleration is zero and the sphere falls at a constant terminal velocity $v_{t}$ .

Left: at terminal velocity the weight (down) is balanced by the upthrust and viscous drag (both up), so the resultant force is zero. Right: the velocity rises with a decreasing gradient (acceleration falls as drag grows) and levels off at the constant terminal velocity.

Writing the three forces for a sphere of radius $r$ (volume $V = \tfrac{4}{3}\pi r^3$ ), density $\rho_{s}$ , in a fluid of density $\rho_{f}$ and viscosity $\eta$ :

$\rho_{s}\tfrac{4}{3}\pi r^3 g = \rho_{f}\tfrac{4}{3}\pi r^3 g + 6\pi\eta r v_{t}.$

The $\pi r$ factors cancel when you rearrange, giving the standard result

$v_{t} = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}.$

Notice the density difference $(\rho_{s} - \rho_{f})$ — that is the upthrust correctly built in.

Core Practical 2 — measuring viscosity by the falling-ball method. Rearranging for $\eta$ ,

$\eta = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9 v_{t}}.$

Drop a steel ball-bearing into a tall measuring cylinder of the liquid.
Mark two lines on the cylinder, the upper one well below the surface so the ball has reached terminal velocity before it passes it (check by timing equal intervals — when the time per interval is constant, the velocity is constant).
Time the ball between the two marks and divide the distance by the time to get $v_{t}$ .
Measure the ball's diameter with a micrometer (at several points, then average and halve to get the radius $r$ ).
Look up or measure $\rho_{s}$ and $\rho_{f}$ , then substitute into the equation above.

Evaluation points examiners reward: because the formula contains $r^2$ , a small radius error is roughly doubled in $\eta$ — measure the diameter carefully with a micrometer. Keep the temperature constant (viscosity is temperature-dependent). Use a wide cylinder to minimise wall effects (which otherwise make $\eta$ come out too high). Repeat and average to reduce timing/reaction-time errors, or use a longer fall distance.

Three forces: weight (down), upthrust (up), viscous drag (up).
At terminal velocity: weight = upthrust + viscous drag, resultant = 0.
$v_{t} = \dfrac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$ — note the density DIFFERENCE.
Core Practical 2: time the ball between marks at terminal velocity; measure $r$ with a micrometer; control temperature.

See the full worked example for density, upthrust and viscous drag →

Quick recap

Density $\rho = m/V$ in $\text{kg m}^{-3}$ ; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ ; average density decides floating/sinking.
Upthrust = weight of fluid displaced (Archimedes): $U = \rho_{f} V g$ , using the fluid's density and submerged volume.
Float if average density < fluid; sink if >; floating fraction submerged $= \rho_{object}/\rho_{fluid}$ .
Viscosity $\eta$ (Pa s) measures resistance to flow; for liquids it falls as temperature rises.
Stokes' law $F = 6\pi\eta r v$ applies to a small smooth sphere in laminar flow (uses the radius).
At terminal velocity weight = upthrust + viscous drag, giving $v_{t} = \dfrac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$ .
Core Practical 2 finds $\eta$ by timing a ball at terminal velocity; control temperature and measure the radius carefully.

Memorise this

Verbatim phrases and definitions Edexcel mark schemes credit.

Density: $\rho = m/V$ , unit $\text{kg m}^{-3}$ ; $1\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ .
Upthrust = weight of fluid displaced; $U = \rho_{f} V g$ (fluid density, submerged volume).
Floats if average density < fluid density; fraction submerged $= \rho_{object}/\rho_{fluid}$ .
Stokes' law: $F = 6\pi\eta r v$ — small smooth sphere, laminar flow, radius not diameter.
Terminal velocity: weight = upthrust + viscous drag, so $v_{t} = \dfrac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$ .
Viscosity of a liquid decreases as temperature increases (gases increase).
Core Practical 2: $\eta = \dfrac{2 r^2 g (\rho_{s} - \rho_{f})}{9 v_{t}}$ — measure $v_t$ between marks, $r$ with a micrometer, control temperature.

How it’s examined

This material sits in Unit 1 (WPH11, Mechanics and Materials) and is tested on the Unit 1 paper through structured questions and Core Practical 2. Expect: a one-step density or unit-conversion calculation (1–2 marks, AO2); an upthrust calculation using $U = \rho_f V g$ (2–3 marks) where the trap is using the object's density; a ‘state and explain whether it floats or sinks’ part needing a density (or weight-vs-upthrust) comparison; a Stokes'-law substitution (watch radius vs diameter); and a terminal-velocity question requiring the full force balance weight = upthrust + viscous drag, often as a ‘show that’ or a derive-and-calculate. Core Practical 2 brings AO3 evaluation marks — how to confirm terminal velocity, why $r$ is measured with a micrometer (the $r^2$ dependence), wall effects, and controlling temperature. Always show the equation, substitution with units, and a 3-s.f. answer to secure the method and accuracy marks.

Sources: Pearson Edexcel International Advanced Level Physics (YPH11/XPH11) specification (Issue 3, July 2021); Edexcel IAL Physics Unit 1 examiner reports; Edexcel IAL Physics Unit 1 sample assessment materials and mark schemes. Last reviewed 2026-06-26.

Take this whole topic with you

Step-by-step worked examples — Density, Upthrust and Viscous Drag

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

Question type:

1Calculating density from mass and volume (2 marks)
Getting startedDirect calculation• density, AO2, Unit 1
Question
A solid aluminium block has a mass of $2.70\ \text{kg}$ and a volume of $1.00 \times 10^{-3}\ \text{m}^3$ . Calculate its density. (2 marks)
Step-by-step solution
1. Step 1
  Select the defining equation for density (mass per unit volume).
  $\rho = \frac{m}{V}$
2. Step 2
  Substitute the values in SI units (mass in kg, volume in m³) and evaluate.
  $\rho = \frac{2.70}{1.00 \times 10^{-3}} = 2700\ \text{kg m}^{-3}$
Answer
$\rho = 2700\ \text{kg m}^{-3}$ (i.e. $2.70 \times 10^3\ \text{kg m}^{-3}$ ).
Examiner tip
Mark scheme: (1) correct substitution into $\rho = m/V$ ; (2) correct answer with unit kg m⁻³. A missing or wrong unit forfeits the accuracy mark. Watch volumes given in cm³ — convert with $1\ \text{cm}^3 = 1 \times 10^{-6}\ \text{m}^3$ .
2Converting g cm⁻³ to kg m⁻³ (2 marks)
Getting startedDirect calculation• density, units, Unit 1
Question
The density of water is often quoted as $1.00\ \text{g cm}^{-3}$ . Show that this is equal to $1000\ \text{kg m}^{-3}$ . (2 marks)
Step-by-step solution
1. Step 1
  Convert the mass unit: $1\ \text{g} = 1 \times 10^{-3}\ \text{kg}$ .
2. Step 2
  Convert the volume unit: $1\ \text{cm}^3 = (1 \times 10^{-2}\ \text{m})^3 = 1 \times 10^{-6}\ \text{m}^3$ , then combine.
  $\rho = \frac{1 \times 10^{-3}\ \text{kg}}{1 \times 10^{-6}\ \text{m}^3} = 1000\ \text{kg m}^{-3}$
Answer
$1.00\ \text{g cm}^{-3} = 1000\ \text{kg m}^{-3}$ (multiply by $10^3$ ).
Examiner tip
‘Show that’ questions need the conversion shown explicitly, not just the answer quoted. The trap is forgetting that the cm → m conversion is cubed (×10⁶), so the overall factor is ×1000, not ×100.
3Upthrust as weight of fluid displaced (3 marks)
Getting startedDirect calculation• upthrust, Archimedes, Unit 1
Question
A metal sphere of volume $2.00 \times 10^{-4}\ \text{m}^3$ is fully submerged in water (density $1000\ \text{kg m}^{-3}$ ). Calculate the upthrust acting on it. Take $g = 9.81\ \text{m s}^{-2}$ . (3 marks)
Step-by-step solution
1. Step 1
  Upthrust equals the weight of the fluid displaced (Archimedes' principle). The volume of fluid displaced equals the submerged volume of the sphere.
  $U = \rho_{f} V g$
2. Step 2
  Substitute the density of water, the volume and g.
  $U = 1000 \times 2.00 \times 10^{-4} \times 9.81$
3. Step 3
  Evaluate and give the answer with its unit.
  $U = 1.96\ \text{N}$
Answer
$U = 1.96\ \text{N}$ (upwards).
Examiner tip
Mark scheme: (1) $U = \rho V g$ / weight of fluid displaced; (2) correct substitution with the density of the FLUID; (3) $1.96\ \text{N}$ . The common error is using the density of the sphere instead of the density of the fluid.
4Will it float? Comparing densities (4 marks)
Building confidenceMulti-step problem• floating, density, AO2, Unit 1
Question
A solid block of mass $0.540\ \text{kg}$ has a volume of $6.00 \times 10^{-4}\ \text{m}^3$ . By calculating its density, determine whether the block floats or sinks in water (density $1000\ \text{kg m}^{-3}$ ), and explain your reasoning. (4 marks)
Step-by-step solution
1. Step 1
  Find the density of the block.
  $\rho = \frac{m}{V} = \frac{0.540}{6.00 \times 10^{-4}} = 900\ \text{kg m}^{-3}$
2. Step 2
  Compare the object's density with the fluid's density. Here $900 < 1000$ , so the block is less dense than water.
3. Step 3
  If fully submerged, the upthrust (weight of an equal volume of water) would exceed the block's weight, so there is a resultant upward force.
4. Step 4
  The block therefore rises until it floats partly out of the water; at equilibrium the upthrust from the SUBMERGED part equals the weight. It floats.
Answer
$\rho_{block} = 900\ \text{kg m}^{-3} < 1000\ \text{kg m}^{-3}$ , so the block floats.
Examiner tip
Mark scheme: (1) density = 900 kg m⁻³; (2) comparison with water's density; (3) reason that upthrust > weight when submerged; (4) conclusion ‘floats’. Stating ‘it floats because it is light’ scores 0 — the comparison of densities (or weight vs upthrust) is required.
5Viscous drag from Stokes' law (3 marks)
Building confidenceDirect calculation• Stokes' law, viscous drag, Unit 1
Question
A smooth steel ball of radius $1.50\ \text{mm}$ moves through oil of viscosity $0.990\ \text{Pa s}$ at a steady speed of $3.50 \times 10^{-3}\ \text{m s}^{-1}$ . The flow is laminar. Calculate the viscous drag force on the ball. (3 marks)
Step-by-step solution
1. Step 1
  Select Stokes' law for the drag on a small sphere in laminar flow. Convert the radius to metres: $r = 1.50 \times 10^{-3}\ \text{m}$ .
  $F = 6\pi\eta r v$
2. Step 2
  Substitute the values in SI units.
  $F = 6\pi \times 0.990 \times (1.50 \times 10^{-3}) \times (3.50 \times 10^{-3})$
3. Step 3
  Evaluate the drag force.
  $F = 9.80 \times 10^{-5}\ \text{N}$
Answer
$F = 9.80 \times 10^{-5}\ \text{N}$ (opposing the motion).
Examiner tip
Mark scheme: (1) Stokes' law selected with radius in metres; (2) correct substitution including the $6\pi$ factor; (3) $9.80 \times 10^{-5}\ \text{N}$ . The classic error is using diameter instead of radius — Stokes' law uses the radius $r$ .
6Terminal velocity of a falling sphere (4 marks)
Building confidenceMulti-step problem• terminal velocity, force balance, AO2, Unit 1
Question
A small steel ball of radius $1.60\ \text{mm}$ and density $7850\ \text{kg m}^{-3}$ falls through glycerol of density $1260\ \text{kg m}^{-3}$ and viscosity $1.50\ \text{Pa s}$ . At terminal velocity, weight = upthrust + viscous drag. Calculate the terminal velocity. ( $g = 9.81\ \text{m s}^{-2}$ , volume of a sphere $= \frac{4}{3}\pi r^3$ .) (4 marks)
Step-by-step solution
1. Step 1
  Write the force balance at terminal velocity, with weight $= \rho_{s}Vg$ and upthrust $= \rho_{f}Vg$ , where $V = \frac{4}{3}\pi r^3$ .
  $\rho_{s}\,\tfrac{4}{3}\pi r^3 g = \rho_{f}\,\tfrac{4}{3}\pi r^3 g + 6\pi\eta r v_{t}$
2. Step 2
  Rearrange for the terminal velocity (the $\pi r$ factors cancel) to give the standard result.
  $v_{t} = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$
3. Step 3
  Substitute the values, with $r = 1.60 \times 10^{-3}\ \text{m}$ .
  $v_{t} = \frac{2 \times (1.60 \times 10^{-3})^2 \times 9.81 \times (7850 - 1260)}{9 \times 1.50}$
4. Step 4
  Evaluate the terminal velocity.
  $v_{t} = 2.45 \times 10^{-2}\ \text{m s}^{-1}$
Answer
$v_{t} = 2.45 \times 10^{-2}\ \text{m s}^{-1}$ (about $2.45\ \text{cm s}^{-1}$ ).
Examiner tip
Mark scheme: (1) correct force balance $W = U + F$ with the upthrust term included; (2) correct rearrangement for $v_t$ ; (3) substitution with $r$ in metres and the density DIFFERENCE; (4) $2.45 \times 10^{-2}\ \text{m s}^{-1}$ . Forgetting upthrust (using $\rho_s$ instead of $\rho_s - \rho_f$ ) is the most penalised slip.
7Core Practical 2 — viscosity from a falling ball (4 marks)
Building confidenceMulti-step problem• Adapted from Edexcel IAL Physics Unit 1 / Core Practical 2• Core Practical, viscosity, AO3, Unit 1
Question
In a falling-ball experiment, a steel ball-bearing of radius $1.50\ \text{mm}$ and density $7800\ \text{kg m}^{-3}$ reaches a terminal velocity of $2.73 \times 10^{-2}\ \text{m s}^{-1}$ in a liquid of density $1260\ \text{kg m}^{-3}$ . Use these data to determine the viscosity of the liquid. ( $g = 9.81\ \text{m s}^{-2}$ .) (4 marks)
Step-by-step solution
1. Step 1
  Start from the terminal-velocity expression derived from the force balance (weight = upthrust + Stokes drag).
  $v_{t} = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$
2. Step 2
  Rearrange to make the viscosity the subject.
  $\eta = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9 v_{t}}$
3. Step 3
  Substitute, with $r = 1.50 \times 10^{-3}\ \text{m}$ .
  $\eta = \frac{2 \times (1.50 \times 10^{-3})^2 \times 9.81 \times (7800 - 1260)}{9 \times (2.73 \times 10^{-2})}$
4. Step 4
  Evaluate, giving the viscosity in pascal seconds.
  $\eta = 1.18\ \text{Pa s}$
Answer
$\eta = 1.18\ \text{Pa s}$ .
Examiner tip
Mark scheme: (1) correct equation including upthrust; (2) rearranged for $\eta$ ; (3) substitution with $r$ in metres and density difference; (4) $1.18\ \text{Pa s}$ (accept 1.17–1.18). Accept the unit written as kg m⁻¹ s⁻¹. Quoting the radius from a measured diameter without halving it is a frequent A-mark loss.
8Fraction of a floating object submerged (5 marks)
StretchMulti-step problem• floating, upthrust, AO2, Unit 1
Question
An iceberg of density $917\ \text{kg m}^{-3}$ floats in sea water of density $1025\ \text{kg m}^{-3}$ . By equating the weight of the iceberg to the upthrust on the submerged part, derive an expression for the fraction of its volume that lies below the surface, and evaluate it. (5 marks)
Step-by-step solution
1. Step 1
  For a floating object, the upthrust equals the weight. Write the weight using the iceberg's total volume $V$ and the upthrust using the submerged volume $V_{sub}$ .
  $\rho_{ice} V g = \rho_{sea} V_{sub} g$
2. Step 2
  Cancel $g$ from both sides — equilibrium is independent of g.
  $\rho_{ice} V = \rho_{sea} V_{sub}$
3. Step 3
  Rearrange to express the submerged fraction as a ratio of densities.
  $\frac{V_{sub}}{V} = \frac{\rho_{ice}}{\rho_{sea}}$
4. Step 4
  Substitute the densities.
  $\frac{V_{sub}}{V} = \frac{917}{1025}$
5. Step 5
  Evaluate the fraction (and express as a percentage).
  $\frac{V_{sub}}{V} = 0.895 \;(89.5\%)$
Answer
Submerged fraction $= \rho_{ice}/\rho_{sea} = 0.895$ , so about $89.5\%$ of the iceberg lies below the surface.
Examiner tip
Mark scheme: (1) upthrust = weight for a floating body; (2)–(3) correct algebra cancelling $g$ and $V$ to reach $V_{sub}/V = \rho_{ice}/\rho_{sea}$ ; (4) substitution; (5) 0.895. A* answers note that $g$ cancels, so the fraction depends only on the density ratio.
9Charged oil drop falling at terminal velocity (5 marks)
StretchMulti-step problem• Adapted from Edexcel IAL Physics Unit 1• Stokes' law, terminal velocity, synoptic, Unit 1
Question
A tiny oil drop of radius $1.00 \times 10^{-6}\ \text{m}$ falls at its terminal velocity through air of viscosity $1.80 \times 10^{-5}\ \text{Pa s}$ . The drop's terminal velocity is measured as $4.00 \times 10^{-5}\ \text{m s}^{-1}$ . Neglecting the upthrust of air, calculate the viscous drag force on the drop and hence estimate its weight. (5 marks)
Step-by-step solution
1. Step 1
  At terminal velocity through air, the upthrust of air is negligible, so weight ≈ viscous drag. Use Stokes' law for the drag.
  $F = 6\pi\eta r v$
2. Step 2
  Substitute the values for the air and the drop.
  $F = 6\pi \times (1.80 \times 10^{-5}) \times (1.00 \times 10^{-6}) \times (4.00 \times 10^{-5})$
3. Step 3
  Evaluate the drag force.
  $F = 1.36 \times 10^{-14}\ \text{N}$
4. Step 4
  Since the air upthrust is neglected and the drop moves at constant velocity, the net force is zero, so the weight equals the drag.
  $W = F = 1.36 \times 10^{-14}\ \text{N}$
5. Step 5
  Note the assumptions: the flow must be laminar and the drop a smooth sphere for Stokes' law to apply, and air upthrust is small because air is far less dense than oil.
Answer
$F = 1.36 \times 10^{-14}\ \text{N}$ ; neglecting air upthrust, the weight $W \approx 1.36 \times 10^{-14}\ \text{N}$ .
Examiner tip
Mark scheme: (1) weight ≈ drag at terminal velocity (air upthrust negligible); (2) correct Stokes substitution; (3) $1.36 \times 10^{-14}\ \text{N}$ ; (4) $W = F$ ; (5) state the laminar-flow / smooth-sphere assumption. This is the Millikan-oil-drop context; examiners credit explicit statement of why upthrust can be ignored.
10Why warmer liquid gives a faster ball (3 marks)
StretchWord problem• viscosity, temperature, A*, Unit 1
Question
In a falling-ball viscosity experiment the temperature of the liquid rose during the afternoon. Explain, in terms of viscosity and the terminal-velocity equation, why the ball reached a higher terminal velocity later in the day, and state how this would affect the calculated value of viscosity if the temperature change were ignored. (3 marks)
Step-by-step solution
1. Step 1
  For a liquid, viscosity DECREASES as temperature increases (the molecules have more kinetic energy and the intermolecular forces are more easily overcome, so layers slide past one another more readily).
2. Step 2
  From $v_{t} = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$ , the terminal velocity is inversely proportional to viscosity, so a lower $\eta$ gives a larger $v_{t}$ — the ball falls faster.
  $v_{t} \propto \frac{1}{\eta}$
3. Step 3
  If the temperature rise is ignored and one viscosity value is quoted, the result is unreliable: the viscosity genuinely changed during the run, so the measured terminal velocities correspond to different $\eta$ values. Temperature must be controlled (e.g. a water bath) for a valid result.
Answer
Higher temperature → lower viscosity → from $v_{t} \propto 1/\eta$ the ball falls faster; ignoring the temperature change makes the viscosity value invalid, so temperature must be controlled.
Examiner tip
Mark scheme: (1) viscosity of a liquid decreases with temperature; (2) link via $v_t \propto 1/\eta$ to a larger terminal velocity; (3) consequence for validity / need to control temperature. A* candidates contrast liquids (viscosity falls) with gases (viscosity rises) when asked generally.

Model Answers — Density, Upthrust and Viscous Drag

High-scoring sample answers for density, upthrust and viscous drag on the Cambridge IGCSE paper, with examiner-style notes mapping each response to the mark scheme and assessment objectives.

Question 1
2 marks
Q (2 marks). Define density and state its SI unit.
Model answer
Density is the mass per unit volume of a material, defined by the equation $\rho = m/V$ , where $m$ is mass and $V$ is volume.

Its SI unit is the kilogram per cubic metre, $\text{kg m}^{-3}$ .
Why this scores
Why this scores 2/2. (1) mass per unit volume / $\rho = m/V$ ; (2) correct unit kg m⁻³. Including the defining equation is the safe way to secure the first mark; ‘how heavy something is’ is not credited.
Question 2
2 marks
Q (2 marks). State Archimedes' principle and explain what is meant by upthrust.
Model answer
Archimedes' principle states that the upthrust on a body immersed (wholly or partly) in a fluid is equal to the weight of the fluid displaced by the body.

Upthrust is the resultant upward force that a fluid exerts on a body within it, arising because the pressure at the bottom of the body is greater than the pressure at the top. It can be calculated from $U = \rho_{f} V g$ , where $\rho_f$ is the fluid density and $V$ the displaced volume.
Why this scores
Why this scores 2/2. (1) upthrust = weight of fluid displaced; (2) upthrust is an upward force / from the pressure difference, with $U = \rho V g$ as supporting detail. Saying ‘the force that makes things float’ is too vague for the second mark.
Question 3
3 marks
Q (3 marks). State Stokes' law and give two conditions that must be met for it to be valid.
Model answer
Stokes' law gives the viscous drag force $F$ on a small sphere moving through a fluid as $F = 6\pi\eta r v$ , where $\eta$ is the viscosity of the fluid, $r$ the radius of the sphere and $v$ its speed relative to the fluid.

For Stokes' law to apply, the flow must be laminar (streamline), not turbulent, and the sphere must be small and smooth, moving slowly through the fluid. (The fluid is also assumed to be of effectively infinite extent — no wall effects.)
Why this scores
Why this scores 3/3. (1) correct statement $F = 6\pi\eta r v$ with terms identified; (2) laminar/streamline flow; (3) small smooth sphere / slow speed (or no wall effects). Two valid conditions earn the final two marks. Writing the equation without identifying the symbols risks losing the first mark.
Question 4
4 marks
Q (4 marks). A steel ball is released from rest just below the surface of a tall column of oil. Describe and explain how the forces acting on the ball change as it falls, until it reaches terminal velocity.
Model answer
Three forces act on the ball: its weight acting downwards, the upthrust acting upwards (the weight of oil displaced), and the viscous drag acting upwards (opposing the motion).

Just after release the ball is slow, so the viscous drag is small ( $F = 6\pi\eta r v$ depends on speed). The weight exceeds the upthrust plus drag, giving a resultant downward force, so the ball accelerates.

As the ball speeds up, the viscous drag increases because it is proportional to speed. The resultant force therefore decreases, so the acceleration decreases.

Eventually the ball is fast enough that weight = upthrust + viscous drag. The resultant force is now zero, the acceleration is zero, and the ball continues at a constant terminal velocity.
Why this scores
Why this scores 4/4. (1) the three forces named with correct directions (weight down; upthrust and drag up); (2) initially drag small → resultant force down → accelerates; (3) drag increases with speed → resultant and acceleration decrease; (4) at terminal velocity weight = upthrust + drag, resultant = 0, constant velocity. Stating drag depends on speed ( $F = 6\pi\eta r v$ ) is the key reasoning link.
Question 5
5 marks
Q (5 marks). A glass sphere of radius $1.00\ \text{mm}$ and density $2500\ \text{kg m}^{-3}$ is dropped into oil of density $920\ \text{kg m}^{-3}$ and viscosity $0.990\ \text{Pa s}$ . (a) Calculate the terminal velocity of the sphere. (b) State one assumption you have made. ( $g = 9.81\ \text{m s}^{-2}$ .)
Model answer
(a) At terminal velocity the resultant force is zero, so weight = upthrust + viscous drag:

$\rho_{s}\tfrac{4}{3}\pi r^3 g = \rho_{f}\tfrac{4}{3}\pi r^3 g + 6\pi\eta r v_{t}$

Rearranging (the $\pi r$ factors cancel) gives

$v_{t} = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}.$

Substituting $r = 1.00 \times 10^{-3}\ \text{m}$ :

$v_{t} = \frac{2 \times (1.00 \times 10^{-3})^2 \times 9.81 \times (2500 - 920)}{9 \times 0.990} = 3.48 \times 10^{-3}\ \text{m s}^{-1}.$

So the terminal velocity is $3.48 \times 10^{-3}\ \text{m s}^{-1}$ (about $3.5\ \text{mm s}^{-1}$ ).

(b) The flow around the sphere is laminar (so Stokes' law applies); the sphere is small and smooth, and the temperature (hence viscosity) is constant.
Why this scores
Why this scores 5/5. (a) 4 marks: correct force balance including upthrust; correct rearrangement; substitution with $r$ in metres and the density DIFFERENCE; answer $3.48 \times 10^{-3}\ \text{m s}^{-1}$ . (b) 1 mark: any valid assumption (laminar flow / smooth small sphere / constant temperature). Using $\rho_s$ rather than $\rho_s - \rho_f$ caps the answer at the method mark.
Question 6
4 marks
Q (4 marks). An alloy is made by mixing two metals so that, by volume, it is $60\%$ copper (density $8960\ \text{kg m}^{-3}$ ) and $40\%$ zinc (density $7140\ \text{kg m}^{-3}$ ). Calculate the average density of the alloy, and explain why this is greater than the density of water.
Model answer
Taking $1\ \text{m}^3$ of alloy, the mass of copper is $0.60 \times 8960 = 5376\ \text{kg}$ and the mass of zinc is $0.40 \times 7140 = 2856\ \text{kg}$ .

The total mass in $1\ \text{m}^3$ is $5376 + 2856 = 8232\ \text{kg}$ , so the average density is

$\rho = \frac{m}{V} = \frac{8232}{1.00} = 8232\ \text{kg m}^{-3} \approx 8230\ \text{kg m}^{-3}.$

This is far greater than the density of water ( $1000\ \text{kg m}^{-3}$ ) because the metal atoms are heavier and more closely packed, so a given volume of alloy contains much more mass than the same volume of water. The alloy therefore sinks in water.
Why this scores
Why this scores 4/4. (1) mass of each metal per unit volume found correctly; (2) total mass and average density calculation; (3) answer ≈ 8230 kg m⁻³; (4) comparison with water density / closer packing → sinks. The average density is the total mass over total volume, NOT the simple mean of the two densities (which would also give 8232 here only because the proportions are by volume).
Question 7
6 marks
Q (6 marks). A student determines the viscosity of a liquid using Core Practical 2 (the falling-ball method). They drop a steel ball-bearing into a tall measuring cylinder of the liquid, mark two lines on the cylinder, and time the ball between them. (a) Explain how the student ensures the ball is travelling at terminal velocity between the two marks. (b) Identify two main sources of error and explain how each affects the result or how to reduce it. (c) State one safety or control measure needed for a valid result.
Model answer
(a) The ball must reach terminal velocity before it passes the first mark, so the upper mark is placed well below the surface — far enough that the ball has stopped accelerating. The student can check this by timing the ball over successive equal intervals: once the time per interval is constant, the velocity is constant, confirming terminal velocity has been reached.

(b) Source 1 — measuring the radius. Stokes' law contains $r^2$ , so a small error in the radius is roughly doubled in the viscosity. The diameter should be measured with a micrometer at several points and averaged, then halved to get the radius. Source 2 — wall effects. If the cylinder is narrow, the liquid near the walls drags on the flow, so Stokes' law (which assumes infinite fluid) overestimates the drag and the calculated viscosity is too high. Using a wide cylinder and a small ball reduces this.

(Other valid errors: timing/reaction-time error in measuring $v_t$ — use a longer fall distance or repeat and average; failure to release the ball centrally.)

(c) The temperature of the liquid must be kept constant (e.g. in a water bath and recorded), because viscosity decreases with temperature; an alternative valid control is repeating the measurement and averaging.
Why this scores
Why this scores 6/6. (a) 2 marks: upper mark below surface so terminal velocity is reached + a way to confirm constant velocity. (b) 2 marks: two genuine error sources with effect/reduction (radius squared → micrometer; wall effects → wide tube; or timing → longer distance/repeat). (c) 1 mark: control temperature (viscosity is temperature-dependent). The sixth mark is awarded for the QUALITY of explanation — linking each error to its effect on $\eta$ rather than just listing it.
Question 8
6 marks
Q (6 marks). A sealed, rigid container of fixed volume $0.0500\ \text{m}^3$ and total mass $40.0\ \text{kg}$ is placed in sea water of density $1025\ \text{kg m}^{-3}$ . (a) Calculate the maximum possible upthrust on the container when it is fully submerged. (b) By comparing this with the weight of the container, determine whether it floats or sinks, and explain in terms of forces. (c) Suggest, using the idea of density, how a submarine is able to control whether it floats or sinks. ( $g = 9.81\ \text{m s}^{-2}$ .)
Model answer
(a) When fully submerged the displaced volume equals the container's volume, so the maximum upthrust is

$U = \rho_{f} V g = 1025 \times 0.0500 \times 9.81 = 503\ \text{N}.$

(b) The weight of the container is $W = mg = 40.0 \times 9.81 = 392\ \text{N}$ . Since the maximum upthrust ( $503\ \text{N}$ ) is greater than the weight ( $392\ \text{N}$ ), there is a resultant upward force when fully submerged. The container therefore rises and floats, settling so that the upthrust from the submerged part exactly equals its weight ( $392\ \text{N}$ ). Equivalently, the average density of the container is $40.0/0.0500 = 800\ \text{kg m}^{-3}$ , which is less than the sea water's $1025\ \text{kg m}^{-3}$ , so it floats.

(c) A submarine changes its average density by flooding or emptying ballast tanks with sea water. Taking in water increases its mass (and hence average density) without changing its volume; when its average density exceeds that of sea water, its weight exceeds the upthrust and it sinks. Expelling the water with compressed air lowers the average density below that of sea water, so the upthrust exceeds the weight and it rises.
Why this scores
Why this scores 6/6. (a) 2 marks: $U = \rho V g$ used correctly with the FLUID density and full submerged volume → 503 N. (b) 2 marks: weight = 392 N and a valid comparison (upthrust > weight → resultant up → floats), or the density-comparison route. (c) 2 marks: submarine alters average density via ballast → density vs sea water determines float/sink. Quoting the average density (800 kg m⁻³) as a cross-check is the kind of synoptic touch that secures the top of the band.
Question 9
Adapted from Edexcel IAL Physics Unit 1 / Core Practical 26 marks
Q (6 marks). A student uses the falling-ball method (Core Practical 2) to find the viscosity of a liquid of density $1260\ \text{kg m}^{-3}$ . They measure the diameter of a steel ball (density $7800\ \text{kg m}^{-3}$ ) as $3.00\ \text{mm} \pm 0.01\ \text{mm}$ , and time the ball falling at terminal velocity over a distance of $0.150\ \text{m} \pm 0.001\ \text{m}$ , taking $5.50\ \text{s} \pm 0.20\ \text{s}$ . (a) Calculate the terminal velocity and hence the viscosity. (b) Estimate the percentage uncertainty in the viscosity, treating the densities as exact. (c) State which measurement contributes most to the uncertainty and suggest how to reduce it. ( $g = 9.81\ \text{m s}^{-2}$ .)
Model answer
(a) The terminal velocity is the distance over the time:

$v_{t} = \frac{0.150}{5.50} = 2.73 \times 10^{-2}\ \text{m s}^{-1}.$

The radius is half the diameter: $r = 1.50\ \text{mm} = 1.50 \times 10^{-3}\ \text{m}$ . From the rearranged terminal-velocity equation,

$\eta = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9 v_{t}} = \frac{2 \times (1.50 \times 10^{-3})^2 \times 9.81 \times (7800 - 1260)}{9 \times (2.73 \times 10^{-2})} = 1.18\ \text{Pa s}.$

(b) The percentage uncertainties add for a product/quotient, and the radius enters as $r^2$ so its percentage uncertainty is doubled:

diameter (= radius): $\dfrac{0.01}{3.00} \times 100 = 0.33\%$ , so $r^2$ contributes $2 \times 0.33\% = 0.67\%$ ;

distance: $\dfrac{0.001}{0.150} \times 100 = 0.67\%$ ;

time: $\dfrac{0.20}{5.50} \times 100 = 3.64\%$ .

Total percentage uncertainty in $\eta = 0.67 + 0.67 + 3.64 = 5.0\%$ (to 2 s.f.). This gives an absolute uncertainty of about $0.05\times 1.18 = 0.06\ \text{Pa s}$ , so $\eta = (1.18 \pm 0.06)\ \text{Pa s}$ .

(c) The time measurement dominates the uncertainty ( $3.64\%$ , mostly from reaction time when starting and stopping the timer). It can be reduced by timing over a longer fall distance (so the time is larger and the reaction-time error is a smaller fraction), by using light gates or a video to remove human reaction time, and by repeating the measurement and averaging.
Why this scores
Why this scores 6/6. (a) 2 marks: correct $v_t = 2.73 \times 10^{-2}\ \text{m s}^{-1}$ and $\eta = 1.18\ \text{Pa s}$ (radius halved, density difference used). (b) 3 marks: each percentage uncertainty correct, the $r$ contribution DOUBLED for $r^2$ , and the three added to ≈ 5%. (c) 1 mark: identify time as the dominant term and a valid improvement (longer distance / light gates / repeat). The doubling of the radius uncertainty because of the $r^2$ term is the discriminator examiners look for at the top of the band.

Key Formulae — Density, Upthrust and Viscous Drag

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

Density
$\rho = \frac{m}{V}$
$\rho$
density / $\text{kg m}^{-3}$
$m$
mass / $\text{kg}$
$V$
volume / $\text{m}^3$
When to use
To find the density of a material, or to find mass/volume given the other two. Convert g cm⁻³ to kg m⁻³ by multiplying by 1000.
Example
Aluminium: $\rho = 2.70 / (1.00 \times 10^{-3}) = 2700\ \text{kg m}^{-3}$ .
Upthrust (Archimedes' principle)
$U = \rho_{f} V g$
$U$
upthrust (buoyancy force) / $\text{N}$
$\rho_{f}$
density of the FLUID / $\text{kg m}^{-3}$
$V$
volume of fluid displaced (submerged volume) / $\text{m}^3$
$g$
gravitational field strength / $\text{N kg}^{-1}$
When to use
Upthrust equals the weight of fluid displaced. Use the fluid's density and the submerged volume. Not in the formulae booklet — recall it.
Example
Sphere $V = 2.0 \times 10^{-4}\ \text{m}^3$ in water: $U = 1000 \times 2.0 \times 10^{-4} \times 9.81 = 1.96\ \text{N}$ .
Stokes' law (viscous drag)
$F = 6\pi\eta r v$
$F$
viscous drag force / $\text{N}$
$\eta$
viscosity (coefficient of viscosity) / $\text{Pa s}$
$r$
radius of the sphere / $\text{m}$
$v$
speed of the sphere relative to the fluid / $\text{m s}^{-1}$
When to use
Drag on a SMALL, SMOOTH sphere moving slowly through a fluid in LAMINAR flow only. Given in the Edexcel formulae booklet (Appendix 7). Uses radius, not diameter.
Example
$F = 6\pi \times 0.990 \times 1.50 \times 10^{-3} \times 3.50 \times 10^{-3} = 9.80 \times 10^{-5}\ \text{N}$ .
Terminal-velocity force balance (falling sphere)
$\rho_{s}\tfrac{4}{3}\pi r^3 g = \rho_{f}\tfrac{4}{3}\pi r^3 g + 6\pi\eta r v_{t}$
$\rho_{s}$
density of the sphere / $\text{kg m}^{-3}$
$\rho_{f}$
density of the fluid / $\text{kg m}^{-3}$
$r$
radius of the sphere / $\text{m}$
$v_{t}$
terminal velocity / $\text{m s}^{-1}$
When to use
At terminal velocity the resultant force is zero: weight = upthrust + viscous drag. Equate the three forces, then rearrange for the unknown.
Example
The three terms are weight (down), upthrust (up) and Stokes drag (up). Set $W = U + F$ and solve.
Terminal velocity (rearranged)
$v_{t} = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9\eta}$
$v_{t}$
terminal velocity / $\text{m s}^{-1}$
$r$
radius of the sphere / $\text{m}$
$\rho_{s} - \rho_{f}$
density difference (sphere − fluid) / $\text{kg m}^{-3}$
$\eta$
viscosity / $\text{Pa s}$
When to use
The result of solving the force balance for a sphere. Rearrange to $\eta = \frac{2 r^2 g(\rho_s - \rho_f)}{9 v_t}$ to find viscosity in Core Practical 2. Must be derived, not in the booklet.
Example
Steel in glycerol: $v_t = \frac{2(1.60\times10^{-3})^2 (9.81)(7850-1260)}{9 \times 1.50} = 2.45\times10^{-2}\ \text{m s}^{-1}$ .

Key Definitions and Keywords — Density, Upthrust and Viscous Drag

Definitions to memorise and the exact keywords mark schemes credit for density, upthrust and viscous drag answers — sharpened from recent examiner reports for the 2026 Cambridge IGCSE sitting.

Density
Examiner keyword
The mass per unit volume of a material, $\rho = m/V$ . SI unit: kg m⁻³. (1 g cm⁻³ = 1000 kg m⁻³.)
Average (mean) density
The total mass of an object divided by its total volume. For a hollow or composite object this can differ greatly from the density of the material it is made of, and it determines whether the whole object floats or sinks.
Upthrust (buoyancy force)
Examiner keyword
The resultant upward force exerted by a fluid on a body immersed in it, arising from the greater fluid pressure on the lower surface than the upper surface. $U = \rho_{f} V g$ .
Archimedes' principle
Examiner keyword
The upthrust on a body wholly or partly immersed in a fluid equals the weight of the fluid displaced by the body.
Fluid
A substance that can flow and takes the shape of its container — a liquid or a gas. Both exert upthrust and viscous drag on objects moving through them.
Floating (condition for)
Examiner keyword
An object floats in equilibrium when the upthrust on the submerged part equals its weight. This occurs when the object's average density is less than the fluid's density.
Sinking (condition for)
An object sinks when its weight exceeds the maximum upthrust (the weight of fluid it could displace) — i.e. when its average density is greater than the fluid's density.
Viscosity (coefficient of viscosity)
Examiner keyword
A measure of a fluid's resistance to flow — how readily one layer of the fluid slides over another. Symbol $\eta$ ; SI unit pascal second, Pa s (equivalently kg m⁻¹ s⁻¹). A more viscous fluid (e.g. honey) flows less easily.
Viscous drag
Examiner keyword
The retarding (frictional) force that a fluid exerts on an object moving through it, acting opposite to the relative motion. For a small sphere in laminar flow it is given by Stokes' law, $F = 6\pi\eta r v$ .
Stokes' law
Examiner keyword
The viscous drag on a small, smooth sphere moving slowly through a fluid in laminar flow: $F = 6\pi\eta r v$ , where $\eta$ is viscosity, $r$ the radius and $v$ the speed relative to the fluid.
Laminar (streamline) flow
Examiner keyword
Flow in which the fluid moves in smooth, parallel layers that do not mix, with no abrupt changes in speed or direction. Stokes' law is valid only for laminar flow.
Turbulent flow
Disordered flow with eddies and mixing between layers, occurring at high speeds or around large/rough objects. Stokes' law does NOT apply to turbulent flow.
Terminal velocity
Examiner keyword
The constant maximum velocity reached by an object falling through a fluid, attained when the resultant force is zero — for a falling sphere when weight = upthrust + viscous drag.
Force balance at terminal velocity
Examiner keyword
For a sphere falling through a fluid at terminal velocity: weight (down) = upthrust (up) + viscous drag (up), so $\rho_{s}\tfrac{4}{3}\pi r^3 g = \rho_{f}\tfrac{4}{3}\pi r^3 g + 6\pi\eta r v_{t}$ .
Falling-ball method (Core Practical 2)
Determining a liquid's viscosity by timing a small sphere falling at terminal velocity between two marks, measuring its radius with a micrometer, and using $\eta = \frac{2 r^2 g (\rho_{s} - \rho_{f})}{9 v_{t}}$ .

Common Mistakes and Misconceptions — Density, Upthrust and Viscous Drag

The traps other students keep falling into on density, upthrust and viscous drag questions — taken from recent Cambridge IGCSE examiner reports and mark schemes — and how to avoid them.

✕Using the density of the OBJECT instead of the density of the FLUID in $U = \rho V g$
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students see one object and one density value and use whatever they are given, without thinking about which density Archimedes' principle requires.
How to avoid it
Upthrust equals the weight of fluid displaced, so always use the density of the fluid with the submerged volume. Write $U = \rho_{f} V g$ with the subscript f as a reminder.
✕Forgetting upthrust at terminal velocity (writing weight = viscous drag instead of weight = upthrust + drag)
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Air-resistance problems from earlier study only balance weight against drag, so students carry that habit into denser fluids where upthrust is significant.
How to avoid it
In a liquid the upthrust is NOT negligible. Always use weight = upthrust + viscous drag, which gives the density DIFFERENCE $(\rho_{s} - \rho_{f})$ in the terminal-velocity formula.
✕Putting the diameter into Stokes' law instead of the radius
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Ball-bearing sizes are quoted as diameters and measured with a micrometer as a diameter, so the value to hand is the diameter.
How to avoid it
Stokes' law uses the radius: halve a measured diameter first. Because the terminal-velocity formula contains $r^2$ , this error changes the result by a factor of four.
✕Leaving the radius in mm or the volume in cm³ when substituting into SI equations
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Data are often given in convenient lab units (mm, cm³, g cm⁻³) and students substitute them straight in.
How to avoid it
Convert everything to SI base units first: mm → m ( $\times 10^{-3}$ ), cm³ → m³ ( $\times 10^{-6}$ ), g cm⁻³ → kg m⁻³ ( $\times 1000$ ). Then the answer comes out in N, m s⁻¹ or Pa s as required.
✕Applying Stokes' law to large/fast objects or turbulent flow
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Students treat $F = 6\pi\eta r v$ as a universal drag formula rather than a special case.
How to avoid it
Stokes' law holds only for a small, smooth sphere in laminar flow. For large or fast objects the flow is turbulent and the formula does not apply — state this assumption in evaluation questions.
✕Treating viscosity as constant and ignoring temperature in the falling-ball experiment
Edexcel IAL Physics Unit 1 examiner reports
Why it happens
Viscosity is quoted as a single number in data books, so students assume it does not change during the experiment.
How to avoid it
Viscosity of a liquid decreases as temperature rises (gases are the opposite), so the temperature must be kept constant and recorded (e.g. a water bath). Quote any viscosity value with the temperature it was measured at.

Density, Upthrust and Viscous Drag — frequently asked questions

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

Density, Upthrust and Viscous Drag

Short Notes - Density, Upthrust and Viscous Drag

Detailed Study Notes

At a glance

What you’ll learn

Quick recap

Memorise this

How it’s examined

Take this whole topic with you

1Calculating density from mass and volume (2 marks)

2Converting g cm⁻³ to kg m⁻³ (2 marks)

3Upthrust as weight of fluid displaced (3 marks)

4Will it float? Comparing densities (4 marks)

5Viscous drag from Stokes' law (3 marks)

6Terminal velocity of a falling sphere (4 marks)

7Core Practical 2 — viscosity from a falling ball (4 marks)

8Fraction of a floating object submerged (5 marks)

9Charged oil drop falling at terminal velocity (5 marks)

10Why warmer liquid gives a faster ball (3 marks)

Density

Upthrust (Archimedes' principle)

Stokes' law (viscous drag)

Terminal-velocity force balance (falling sphere)

Terminal velocity (rearranged)

Density

Average (mean) density

Upthrust (buoyancy force)

Archimedes' principle

Fluid

Floating (condition for)

Sinking (condition for)

Viscosity (coefficient of viscosity)

Viscous drag

Stokes' law

Laminar (streamline) flow

Turbulent flow

Terminal velocity

Force balance at terminal velocity

Falling-ball method (Core Practical 2)

✕Using the density of the OBJECT instead of the density of the FLUID in U=ρVgU = \rho V gU=ρVg

✕Forgetting upthrust at terminal velocity (writing weight = viscous drag instead of weight = upthrust + drag)

✕Putting the diameter into Stokes' law instead of the radius

✕Leaving the radius in mm or the volume in cm³ when substituting into SI equations

✕Applying Stokes' law to large/fast objects or turbulent flow

✕Treating viscosity as constant and ignoring temperature in the falling-ball experiment

Density, Upthrust and Viscous Drag — frequently asked questions

✕Using the density of the OBJECT instead of the density of the FLUID in $U = \rho V g$