ExaminerPrep › Question bank › IAL Physics › 1.7 Density, upthrust and viscous drag
Edexcel International A Level Physics · Unit 1 · Mechanics and Materials
40 free, high-quality past-paper-style questions on Density, upthrust and viscous drag for Edexcel IAL Physics, each with a full mark scheme. For when you have finished the past papers, or want to practise one topic at a time.IAL Physics notes and flashcards →Not sure of your level? Diagnostic test →Multiple choice or structured questions, with the mark scheme as a separate PDF.
A souvenir coin has a mass of 8.50 g and a volume of 0.960 cm3.
What is the density of the metal of the coin?
| Answer | Mark | Notes |
|---|---|---|
| ρ = 8.50 ÷ 0.960 = 8.85 g cm−3. Since 1 g cm−3 = 1000 kg m−3, ρ = 8.85 × 103 kg m−3. | ✓ 1 | Answer C |
Answer: C
Common error: 1 m3 = 106 cm3 but 1 kg = 103 g, so g cm−3 → kg m−3 is × 1000, not × 106.
A sealed, empty plastic bottle floats at rest on a pond with part of it above the water.
Which statement about the upthrust on the bottle is correct?
| Answer | Mark | Notes |
|---|---|---|
| The bottle is at rest, so the resultant force is zero: upthrust = weight. It floats just deep enough to displace its own weight of water. | ✓ 1 | Answer D |
Answer: D
Common error: Using the whole volume of a floating body for the upthrust. Only the volume below the surface displaces water.
A steel ball falls at terminal velocity through a tall tube of oil. The oil is then warmed by 10 °C and the ball is dropped again. The densities hardly change.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| The viscosity of a liquid falls as its temperature rises. The drag at any speed is smaller, so the ball must fall faster before the drag balances weight − upthrust: the terminal velocity increases because the viscosity decreases. | ✓ 1 | Answer A |
Answer: A
Common error: Thinking warmer liquids are “thicker”. For liquids, viscosity decreases with temperature — warm syrup pours easily.
A hollow plastic float in an ornamental pond is held completely under the water by a vertical cord tied to the bottom of the pond. The float has a mass of 0.060 kg and a volume of 4.2 × 10−4 m3. Density of water = 1000 kg m−3.
Explain why the water exerts an upward force on the float.
Calculate the upthrust on the float.
Calculate the tension in the cord.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the pressure in the water increases with depth | ✓ 1 | |
| so the upward force on the lower surface is greater than the downward force on the upper surface, giving a resultant force upwards | ✓ 1 | |
| Part (b) | ||
| U = 1000 × 4.2 × 10−4 × 9.81 = 4.1 N (4.12 N) | ✓ 1 | |
| Part (c) | ||
| weight = 0.060 × 9.81 = 0.589 N; equilibrium: T + W = U | ✓ 1 | |
| T = 4.12 − 0.589 = 3.5 N | ✓ 1 | |
Common error: Putting the tension on the same side as the upthrust. Here the cord pulls the float down, so tension + weight = upthrust.
A student makes a simple hydrometer from a plastic drinking straw sealed at the bottom and weighted with modelling clay so that it floats upright. Treat it as a uniform cylinder of cross-sectional area 0.80 cm2 and total mass 6.0 g.
Explain why the hydrometer floats higher in a denser liquid.
Calculate the length of straw below the surface when it floats in water of density 1.00 g cm−3.
In a salt solution, 6.5 cm of the straw is below the surface. Calculate the density of the salt solution in kg m−3.
Suggest one change to the straw that would make the hydrometer more sensitive.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| it floats when the weight of liquid displaced equals its own (fixed) weight | ✓ 1 | |
| in a denser liquid a smaller volume weighs the same, so less of the straw is submerged | ✓ 1 | |
| Part (b) | ||
| 6.0 ÷ (1.00 × 0.80) = 7.5 cm | ✓ 1 | |
| Part (c) | ||
| volume displaced = 0.80 × 6.5 = 5.2 cm3 | ✓ 1 | |
| ρ = 6.0 ÷ 5.2 = 1.15 g cm−3 = 1.15 × 103 kg m−3 | ✓ 1 | |
| Part (d) | ||
| use a narrower straw (smaller cross-sectional area), so a given change in density gives a larger change in the submerged length | ✓ 1 | |
Exam tip: Work in grams and cm3 throughout, then convert the final density: 1 g cm−3 = 1000 kg m−3.
A laboratory compares an engine oil at two temperatures. The same small steel ball is dropped into the oil each time. At 20 °C its terminal velocity is 4.2 mm s−1; at 60 °C it is 31 mm s−1. Stokes' law applies, and the change in the density of the oil can be ignored.
Describe the difference between laminar flow and turbulent flow.
Calculate (viscosity of the oil at 20 °C) ÷ (viscosity of the oil at 60 °C).
Explain why the temperature of the liquid must be kept constant during a falling-ball experiment.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| laminar: the fluid moves in smooth layers that do not mix; the velocity at any point is constant | ✓ 1 | |
| turbulent: the fluid swirls in eddies and mixes; the velocity at a point changes irregularly | ✓ 1 | |
| Part (b) | ||
| the weight minus upthrust is the same, so 6πηrv is the same: η ∝ 1/v | ✓ 1 | |
| ratio = 31 ÷ 4.2 = 7.4 | ✓ 1 | |
| Part (c) | ||
| viscosity depends strongly on temperature, so a changing temperature would change the terminal velocity during the experiment | ✓ 1 | |
Exam tip: In a ratio question, write the equation for each case and divide: everything that is the same cancels.
The graph shows how the mass m of samples of two materials, P and Q, varies with their volume V.
What is the density of material Q?
| Answer | Mark | Notes |
|---|---|---|
| Density = gradient of the line for Q = 200 g ÷ 80 cm3 = 2.5 g cm−3. Since 1 g cm−3 = 1000 kg m−3, ρ = 2500 kg m−3. | ✓ 1 | Answer C |
Answer: C
Exam tip: On a mass–volume graph the density is the gradient. Read a point far along the line, such as (80, 200), so the reading is precise.
The diagram shows a block of wood of density 600 kg m−3 floating in water of density 1000 kg m−3. A metal weight placed on top of the block makes the top of the block exactly level with the water surface; the weight itself is above the water.
What is the mass of the metal weight?
| Answer | Mark | Notes |
|---|---|---|
| Volume of block = 20.0 × 10.0 × 5.0 = 1000 cm3 = 1.0 × 10−3 m3, all under water. Mass of water displaced = 1000 × 1.0 × 10−3 = 1.0 kg = total mass floating. Mass of wood = 600 × 1.0 × 10−3 = 0.60 kg, so the weight has mass 1.0 − 0.60 = 0.40 kg. | ✓ 1 | Answer A |
Answer: A
Exam tip: For a floating system, total mass = mass of fluid displaced. Include everything that is supported — here the block and the weight on it.
The diagrams show the flow of a liquid past two spheres, X and Y.
Which statement is correct?
| Answer | Mark | Notes |
|---|---|---|
| In X the streamlines are smooth, do not cross and the layers do not mix: laminar flow. In Y the fluid forms eddies behind the sphere: turbulent flow. Stokes' law applies only to laminar flow (small spheres, low speed), so X only. | ✓ 1 | Answer A |
Answer: A
Exam tip: Laminar: smooth layers, streamlines never cross, velocity at a point constant. Turbulent: eddies and mixing. Stokes' law needs laminar flow.
The diagram shows a rectangular block and a solid cylinder made of the same metal. The mass of the block is 0.405 kg.
Calculate the volume of the block in m3.
Calculate the density of the metal in kg m−3.
Calculate the mass of the cylinder.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = 12.0 × 5.0 × 2.5 = 150 cm3 = 1.50 × 10−4 m3 | ✓ 1 | |
| Part (b) | ||
| ρ = m ÷ V = 0.405 ÷ 1.50 × 10−4 | ✓ 1 | |
| = 2.70 × 103 kg m−3 | ✓ 1 | |
| Part (c) | ||
| V = π × (2.0 × 10−2)2 × 6.0 × 10−2 = 7.54 × 10−5 m3 | ✓ 1 | |
| m = 2.70 × 103 × 7.54 × 10−5 = 0.204 kg | ✓ 1 | |
Common error: Using the diameter as the radius in πr2h. Halve the diameter first: 4.0 cm → 2.0 cm.
A diver recovers a sealed metal tool box of mass 60 kg from the sea bed by tying it to a lifting bag and filling the bag with air, as shown in the diagram. Ignore the mass of the bag and of the air in it.
Calculate the upthrust on the box.
Calculate the smallest volume of air in the bag that will lift the box off the sea bed.
Once the box leaves the sea bed, no air is added to or removed from the bag. Ignoring drag, explain why the resultant upward force on the box and bag increases as they rise.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = 0.40 × 0.25 × 0.20 = 0.020 m3 | ✓ 1 | |
| U = 1030 × 0.020 × 9.81 = 202 N | ✓ 1 | |
| Part (b) | ||
| weight − upthrust on box = 60 × 9.81 − 202 = 387 N, which the bag's upthrust must supply | ✓ 1 | |
| V = 387 ÷ (1030 × 9.81) = 0.038 m3 | ✓ 1 | |
| Part (c) | ||
| the water pressure decreases as they rise, so the air in the bag expands | ✓ 1 | |
| the bag displaces more water, so the upthrust and therefore the resultant upward force increase | ✓ 1 | |
Exam tip: Treat the box and bag as one system: total upthrust must at least equal total weight. The box's own upthrust reduces the help the bag must give.
A student pours a liquid into a measuring cylinder on a balance, 20 cm3 at a time, and records the total mass m of the cylinder and liquid against the volume V of liquid. The graph shows the results.
State what the intercept on the m axis represents, and give its value.
Use the graph to determine the density of the liquid in kg m−3.
The experiment is repeated with the same cylinder and a liquid of density 1.20 g cm−3. State how its line would compare with this one.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the mass of the empty measuring cylinder | ✓ 1 | |
| 120 g | ✓ 1 | |
| Part (b) | ||
| density = gradient; use a large triangle, e.g. from (0, 120) to (100, 200) | ✓ 1 | |
| gradient = 80 g ÷ 100 cm3 = 0.80 g cm−3 | ✓ 1 | |
| = 8.0 × 102 kg m−3 (800 kg m−3) | ✓ 1 | |
| Part (c) | ||
| same intercept (120 g) but steeper, with gradient 1.20 g cm−3, i.e. 1.5 times the gradient | ✓ 1 | |
Common error: Dividing a total mass by a volume read off the graph, e.g. 200 g ÷ 100 cm3. That includes the cylinder's mass; the gradient does not.
A lead fishing weight has a weight of 6.0 N and a volume of 5.4 × 10−5 m3. It hangs at rest on a vertical line, completely under water in the sea. Density of sea water = 1030 kg m−3.
What is the tension in the line?
| Answer | Mark | Notes |
|---|---|---|
| Upthrust = 1030 × 5.4 × 10−5 × 9.81 = 0.546 N. In equilibrium, T + U = W, so T = 6.0 − 0.546 = 5.5 N. | ✓ 1 | Answer B |
Answer: B
Exam tip: For a hanging body in a fluid, draw three forces: tension and upthrust up, weight down. The tension is the weight minus the upthrust.
Stokes' law gives the viscous drag F on a sphere of radius r moving at speed v through a fluid of viscosity η as F = 6πηrv.
Which of these is the unit of viscosity expressed in SI base units?
| Answer | Mark | Notes |
|---|---|---|
| η = F/(6πrv). Units: kg m s−2 ÷ (m × m s−1) = kg m−1 s−1 (the same as Pa s). | ✓ 1 | Answer A |
Answer: A
Exam tip: Write the newton as kg m s−2 first, then cancel powers of m and s one at a time. 6π has no unit.
A pollen grain can be modelled as a sphere of diameter 30 μm. It falls through still air at a steady 0.030 m s−1. Viscosity of air = 1.8 × 10−5 Pa s.
What is the viscous drag on the grain?
| Answer | Mark | Notes |
|---|---|---|
| Radius = 15 μm = 1.5 × 10−5 m. F = 6π × 1.8 × 10−5 × 1.5 × 10−5 × 0.030 = 1.5 × 10−10 N. | ✓ 1 | Answer C |
Answer: C
Common error: Substituting the diameter into Stokes' law. The formula uses the radius, so a diameter doubles the answer.
A solid plastic cube floats at rest in a sugar solution of density 1250 kg m−3, with 0.72 of its volume below the surface.
What is the density of the plastic?
| Answer | Mark | Notes |
|---|---|---|
| Floating: weight = upthrust, so ρpVg = 1250 × 0.72Vg. ρp = 1250 × 0.72 = 900 kg m−3. | ✓ 1 | Answer B |
Answer: B
Exam tip: For a floating body, fraction submerged = density of body ÷ density of fluid. A body that floats must be less dense than the fluid.
In a falling-ball experiment, a student uses a stopwatch with a resolution of 0.01 s to time a ball between two marks. Her four readings are 4.82 s, 4.95 s, 4.71 s and 4.88 s.
What is the percentage uncertainty in the mean time?
| Answer | Mark | Notes |
|---|---|---|
| Mean = 19.36 ÷ 4 = 4.84 s. The readings vary far more than the resolution, so uncertainty = half the range = (4.95 − 4.71) ÷ 2 = 0.12 s. Percentage uncertainty = 0.12 ÷ 4.84 × 100 = 2.5%. | ✓ 1 | Answer D |
Answer: D
Exam tip: When repeat readings spread by more than the instrument resolution, the spread (half the range) sets the uncertainty, not the resolution.
A student finds the density of a small stone. She fills a displacement (eureka) can with water until it overflows, waits for the dripping to stop, then lowers the stone in on a thread. The water that overflows is collected in a beaker on a top-pan balance with a resolution of 0.1 g. The mass of water collected is 18.6 g. The mass of the stone is 49.5 g. Density of water = 1.00 g cm−3.
Explain how these measurements give the volume of the stone.
Calculate the density of the stone in kg m−3.
The overflow could instead have been collected in a measuring cylinder with 1 cm3 graduations. Calculate the percentage uncertainty in the volume measured this way, and state why weighing the water is the better method.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the stone displaces its own volume of water, which overflows into the beaker | ✓ 1 | |
| volume = mass of water ÷ density of water = 18.6 g ÷ 1.00 g cm−3 = 18.6 cm3 | ✓ 1 | |
| Part (b) | ||
| ρ = 49.5 ÷ 18.6 = 2.66 g cm−3 | ✓ 1 | |
| = 2.66 × 103 kg m−3 | ✓ 1 | |
| Part (c) | ||
| uncertainty = half a division = 0.5 cm3, so 0.5 ÷ 18.6 × 100 = 2.7% (allow ±1 cm3 → 5.4%) | ✓ 1 | |
| the balance gives 0.05 ÷ 18.6 × 100 = 0.27%, a percentage uncertainty about ten times smaller | ✓ 1 | |
Exam tip: With a displacement can, the mass of the overflow divided by the density of water gives the volume directly. Weighing is usually more precise than reading a cylinder.
A river barge has a flat bottom and vertical sides; its base is a rectangle 18.0 m long and 5.0 m wide. The empty barge has a mass of 4.5 × 104 kg. Density of river water = 1000 kg m−3.
State the condition for the barge to float at rest.
Calculate the depth of the base of the empty barge below the water surface.
The barge is loaded with 2.7 × 104 kg of gravel. Calculate the new depth.
The loaded barge travels out to sea, where the water has a density of 1025 kg m−3. Explain whether it floats higher or lower, and calculate the new depth.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the upthrust equals the weight of the barge / the weight of water displaced equals the weight of the barge | ✓ 1 | |
| Part (b) | ||
| mass of water displaced = 4.5 × 104 kg, so 1000 × (18.0 × 5.0) × h = 4.5 × 104 | ✓ 1 | |
| h = 0.50 m | ✓ 1 | |
| Part (c) | ||
| h = 7.2 × 104 ÷ (1000 × 90) = 0.80 m | ✓ 1 | |
| Part (d) | ||
| higher: sea water is denser, so a smaller volume must be displaced to give the same upthrust | ✓ 1 | |
| h = 7.2 × 104 ÷ (1025 × 90) = 0.78 m | ✓ 1 | |
Exam tip: For a floating body, write mass of fluid displaced = mass of body; then g cancels and the volume displaced follows directly.
A fog droplet is a sphere of water of radius 8.0 μm, falling through still air. Density of water = 1000 kg m−3; viscosity of air = 1.8 × 10−5 Pa s; density of air = 1.2 kg m−3.
State two conditions needed for Stokes' law to give the drag on the droplet.
Show that the weight of the droplet is about 2.1 × 10−11 N.
Calculate the terminal velocity of the droplet. Ignore upthrust.
Explain why it is reasonable to ignore the upthrust on the droplet.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| any two from: the droplet is a small sphere; it moves at low speed; the flow around it is laminar (no turbulence) | — | |
| first condition | ✓ 1 | |
| second condition | ✓ 1 | |
| Part (b) | ||
| V = 4/3π(8.0 × 10−6)3 = 2.14 × 10−15 m3 | ✓ 1 | |
| W = 1000 × 2.14 × 10−15 × 9.81 = 2.10 × 10−11 N | ✓ 1 | |
| Part (c) | ||
| 6πηrv = W, so v = 2.10 × 10−11 ÷ (6π × 1.8 × 10−5 × 8.0 × 10−6) | ✓ 1 | |
| v = 7.8 × 10−3 m s−1 (7.75 × 10−3) | ✓ 1 | |
| Part (d) | ||
| air is about 800 times less dense than water, so the upthrust is only about 0.1% of the weight | ✓ 1 | |
Exam tip: In a “show that” question, give your answer to one more significant figure than the value quoted — 2.10 × 10−11 N here.
A student measures the viscosity of a silicone oil by the falling-ball method. A steel ball of diameter 3.00 mm (measured with a micrometer of resolution 0.01 mm) is dropped into a tall tube of the oil. Four marks, A, B, C and D, are 0.150 m apart, with A well below the surface. The times taken are: A to B 2.25 s, B to C 2.23 s, C to D 2.24 s. Density of steel = 7800 kg m−3; density of the oil = 970 kg m−3.
Explain how the times show that the ball had reached terminal velocity before A.
Calculate the terminal velocity of the ball.
Calculate the viscosity of the oil.
Calculate the percentage uncertainty in the diameter of the ball.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| the ball takes the same time (within experimental variation) to cover each equal distance, so its speed is constant | ✓ 1 | |
| Part (b) | ||
| total time A to D = 6.72 s for 0.450 m | ✓ 1 | |
| v = 0.450 ÷ 6.72 = 0.0670 m s−1 | ✓ 1 | |
| Part (c) | ||
| weight = upthrust + drag: 4/3πr3(ρs − ρf)g = 6πηrv | ✓ 1 | |
| η = 2r2g(ρs − ρf) ÷ 9v = 2 × (1.50 × 10−3)2 × 9.81 × 6830 ÷ (9 × 0.0670) | ✓ 1 | |
| η = 0.500 Pa s | ✓ 1 | |
| Part (d) | ||
| 0.005 ÷ 3.00 × 100 = 0.17% (allow 0.01 ÷ 3.00 × 100 = 0.33%) | ✓ 1 | |
Common error: Using the diameter, 3.00 mm, as r. The radius is 1.50 mm, and because r is squared the error would be a factor of four.
A uniform wooden rod with a small weight at its lower end floats upright, first in water and then in liquid X, as shown. Density of water = 1000 kg m−3.
What is the density of liquid X?
| Answer | Mark | Notes |
|---|---|---|
| The rod floats in both, so it displaces the same mass of liquid each time: ρX × A × 15 = 1000 × A × 12. ρX = 1000 × 12 ÷ 15 = 800 kg m−3. | ✓ 1 | Answer B |
Answer: B
Common error: Inverting the ratio. The rod sinks deeper in liquid X, so X must be less dense than water.
A small steel ball is released from rest in a tall tube of oil. The graph shows how its velocity v changes with time t. Stokes' law applies throughout.
At which point is the acceleration of the ball half of its acceleration at the moment of release?
| Answer | Mark | Notes |
|---|---|---|
| Resultant force = W − U − F, with W and U constant and drag F ∝ v. At release F = 0; at terminal velocity (40 mm s−1) F = W − U. The resultant, and so the acceleration, is halved when F = ½(W − U), i.e. when v = 20 mm s−1: point Q. | ✓ 1 | Answer B |
Answer: B
Exam tip: Because Stokes' drag is proportional to speed, the acceleration falls in proportion to (terminal velocity − v). Halfway to terminal velocity means half the initial acceleration.
A small air bubble rises at a constant velocity through a glass of syrup. The weight of the air in the bubble is negligible.
Which diagram shows the forces acting on the bubble? The lengths of the arrows represent the sizes of the forces.
| Answer | Mark | Notes |
|---|---|---|
| The bubble moves up, so the viscous drag acts down, opposing the motion. With the weight negligible, only upthrust (up) and drag (down) act, and at constant velocity they must be equal: diagram C. | ✓ 1 | Answer C |
Answer: C
Common error: Drawing drag upwards out of habit. Drag always opposes the velocity — for a rising bubble it points down.
A student finds the density of some glass marbles. The diagram shows the water level in a measuring cylinder before and after the marbles are added, and the mass of the marbles.
What is the density of the glass?
| Answer | Mark | Notes |
|---|---|---|
| Volume of marbles = 68 − 42 = 26 cm3. ρ = 65.0 ÷ 26 = 2.50 g cm−3. | ✓ 1 | Answer D |
Answer: D
Common error: Dividing by one of the readings instead of the change in reading. The marbles' volume is the rise in the level.
In a falling-ball experiment, a student starts a stopwatch as she releases a ball at the surface of a liquid. A reference band R is 5.0 cm below the surface. She records the time t at which the ball passes further bands at distances d below R. The ball reaches its terminal velocity almost immediately. The graph shows her results.
What is the terminal velocity of the ball?
| Answer | Mark | Notes |
|---|---|---|
| Velocity = 1 ÷ gradient. Gradient = (11.0 − 3.0) s ÷ (50 − 10) cm = 0.20 s cm−1, so v = 5.0 cm s−1 = 0.050 m s−1. (The 1.0 s intercept is the time taken to fall the 5.0 cm to band R.) | ✓ 1 | Answer C |
Answer: C
Common error: Dividing one distance by one time when the graph has an intercept. Use the gradient, which removes the constant offset.
A small raft for a lake is made from a wooden deck fixed on top of four identical sealed plastic drums, arranged in two rows of two. The diagram shows the dimensions of a drum. The total mass of the raft (deck and drums) is 120 kg. Density of fresh water = 1000 kg m−3.
Calculate the volume of one drum.
Calculate the volume of water displaced by the unloaded raft, and hence the fraction of each drum that is under water.
Calculate the greatest load the raft can carry in fresh water before the drums are completely submerged.
The raft is moved to a salt lake where the water has a density of 1180 kg m−3. Calculate the new greatest load.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = π × 0.282 × 0.90 = 0.222 m3 | ✓ 1 | |
| Part (b) | ||
| volume displaced = 120 ÷ 1000 = 0.120 m3 | ✓ 1 | |
| per drum 0.030 m3, fraction = 0.030 ÷ 0.222 = 0.14 | ✓ 1 | |
| Part (c) | ||
| maximum upthrust = weight of 4 × 0.222 m3 of water: mass 887 kg | ✓ 1 | |
| load = 887 − 120 = 770 kg (767 kg) | ✓ 1 | |
| Part (d) | ||
| mass of salt water displaced when fully submerged = 1180 × 0.887 = 1046 kg | ✓ 1 | |
| load = 1046 − 120 = 930 kg (926 kg) | ✓ 1 | |
Exam tip: The greatest load is reached when the whole floating volume is under water: (mass of fluid in that volume) − (mass of the raft).
A small steel ball is released from rest in a deep tank of glycerol. The free-body diagram shows the forces on the ball at one instant, when its speed is 0.040 m s−1. Stokes' law applies.
Calculate the acceleration of the ball at this instant.
Calculate the terminal velocity of the ball.
Explain why the upthrust on the ball stays the same as it falls but the drag does not.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| resultant = 3.60 − 0.58 − 1.80 = 1.22 mN; mass = 3.60 × 10−3 ÷ 9.81 = 3.67 × 10−4 kg | ✓ 1 | |
| a = 1.22 × 10−3 ÷ 3.67 × 10−4 = 3.3 m s−2 | ✓ 1 | |
| Part (b) | ||
| at terminal velocity drag = weight − upthrust = 3.02 mN | ✓ 1 | |
| Stokes' law: drag ∝ speed | ✓ 1 | |
| v = 0.040 × 3.02 ÷ 1.80 = 0.067 m s−1 | ✓ 1 | |
| Part (c) | ||
| upthrust = weight of glycerol displaced, and the ball's volume and the glycerol's density do not change | ✓ 1 | |
| drag depends on speed (F = 6πηrv), and the speed increases until terminal velocity is reached | ✓ 1 | |
Exam tip: With drag proportional to speed, a ratio is quicker than finding η and r: terminal speed = present speed × (W − U) ÷ present drag.
A student uses balls that all have radius 1.00 mm but are made of six different materials, from glass to lead. She measures the terminal velocity v of each ball in the same tube of glycerol, of density 1260 kg m−3, kept at a constant temperature. The graph shows v against (ρball − ρliquid) with a line of best fit.
Show that v should be proportional to (ρball − ρliquid) when Stokes' law applies.
Determine the gradient of the line in SI units.
Calculate the viscosity of the glycerol.
Suggest how the student could keep the temperature of the glycerol constant, and why this matters.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at terminal velocity weight = upthrust + drag: 4/3πr3ρballg = 4/3πr3ρliquidg + 6πηrv | ✓ 1 | |
| v = (2r2g ÷ 9η) × (ρball − ρliquid), and the bracket is constant because r, g and η are the same for every ball | ✓ 1 | |
| Part (b) | ||
| gradient = 12.0 mm s−1 ÷ 8.0 × 103 kg m−3 | ✓ 1 | |
| = 12.0 × 10−3 ÷ 8.0 × 103 = 1.5 × 10−6 m4 kg−1 s−1 (allow 1.45–1.55) | ✓ 1 | |
| Part (c) | ||
| η = 2r2g ÷ (9 × gradient) = 2 × (1.00 × 10−3)2 × 9.81 ÷ (9 × 1.5 × 10−6) | ✓ 1 | |
| η = 1.45 Pa s (allow 1.4–1.5) | ✓ 1 | |
| Part (d) | ||
| stand the tube in a water bath / leave it to reach room temperature and monitor with a thermometer, because viscosity changes with temperature | ✓ 1 | |
Exam tip: Plot the variable you changed against the one you measured so that the constant terms form the gradient: here gradient = 2r2g/9η.
A solid metal cylinder of cross-sectional area 15.0 cm2 hangs upright from a newtonmeter. It is lowered slowly into a beaker of liquid, and the newtonmeter reading is recorded against the depth of the bottom of the cylinder below the liquid surface. The graph shows the results.
Explain why the reading decreases as the depth increases from 0 to 8.0 cm.
Explain why the graph becomes horizontal.
Use the graph to find the height of the cylinder and the upthrust when it is fully submerged, and calculate the density of the liquid.
Calculate the density of the metal.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| a larger volume of the cylinder is below the surface, so it displaces more liquid and the upthrust increases | ✓ 1 | |
| reading = weight − upthrust, so the reading falls | ✓ 1 | |
| Part (b) | ||
| the cylinder is completely submerged, so the volume displaced, and hence the upthrust, no longer changes with depth | ✓ 1 | |
| Part (c) | ||
| height = 8.0 cm, so V = 15.0 × 8.0 = 120 cm3 = 1.20 × 10−4 m3 | ✓ 1 | |
| upthrust = 4.00 − 2.80 = 1.20 N | ✓ 1 | |
| ρ = 1.20 ÷ (1.20 × 10−4 × 9.81) = 1.02 × 103 kg m−3 | ✓ 1 | |
| Part (d) | ||
| mass = 4.00 ÷ 9.81 = 0.408 kg | ✓ 1 | |
| ρ = 0.408 ÷ 1.20 × 10−4 = 3.40 × 103 kg m−3 | ✓ 1 | |
Exam tip: The kink in the graph tells you the height of the cylinder: the reading stops changing at the depth where the top goes under.
Two solid spheres, P and Q, are made of the same steel. The radius of Q is twice the radius of P. Each falls at terminal velocity through the same oil, and Stokes' law applies to both.
What is (terminal velocity of Q) ÷ (terminal velocity of P)?
| Answer | Mark | Notes |
|---|---|---|
| At terminal velocity, (ρs − ρf) × 4/3πr3g = 6πηrv, so v ∝ r2. Doubling r multiplies v by 4. | ✓ 1 | Answer C |
Answer: C
Exam tip: For proportionality questions, write the full balance of forces, cancel everything that is the same, and see what power of r is left.
A weather balloon is filled with 3.5 m3 of helium of density 0.18 kg m−3. The envelope and instruments have a total mass of 2.6 kg; their volume is negligible. Density of the surrounding air = 1.20 kg m−3.
What is the resultant force on the balloon just after it is released?
| Answer | Mark | Notes |
|---|---|---|
| Upthrust = 1.20 × 3.5 × 9.81 = 41.2 N. Weight = (0.18 × 3.5 + 2.6) × 9.81 = 31.7 N. Resultant = 41.2 − 31.7 = 9.5 N upwards. | ✓ 1 | Answer B |
Answer: B
Common error: Forgetting that the gas inside the balloon has weight too. The upthrust depends on the air displaced; the weight includes the helium.
A hot-air balloon has an envelope of volume 3200 m3. The envelope, basket, burner and passengers have a total mass of 780 kg; their volume can be ignored. On the morning of the flight, the outside air has a density of 1.18 kg m−3.
Calculate the upthrust on the balloon.
Calculate the greatest density the hot air inside the envelope can have if the balloon is to lift off.
Later in the day the outside air is warmer. Explain why the pilot has to heat the air in the envelope to a higher temperature to lift the same load.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| upthrust = weight of outside air displaced | ✓ 1 | |
| = 1.18 × 3200 × 9.81 = 3.70 × 104 N | ✓ 1 | |
| Part (b) | ||
| lift-off when ρhot × 3200 + 780 ≤ 1.18 × 3200 | ✓ 1 | |
| ρhot = 1.18 − 780 ÷ 3200 = 0.936 kg m−3 | ✓ 1 | |
| Part (c) | ||
| warmer outside air is less dense, so the upthrust (weight of air displaced) is smaller | ✓ 1 | |
| so the air inside must be made even less dense, i.e. hotter, to keep the total weight no greater than the upthrust | ✓ 1 | |
Common error: Comparing the upthrust with the weight of the envelope and passengers only. The hot air inside is part of what must be lifted.
A small glass bead of radius 1.5 mm is released from rest just below the surface of a deep jar of honey. Density of glass = 2500 kg m−3; density of honey = 1420 kg m−3.
Calculate the upthrust on the bead.
Explain the motion of the bead from the moment it is released until it is moving at a steady speed. Then explain how this steady speed would be different if the honey were warmer.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| V = 4/3π(1.5 × 10−3)3 = 1.41 × 10−8 m3 | ✓ 1 | |
| U = 1420 × 1.41 × 10−8 × 9.81 = 2.0 × 10−4 N | ✓ 1 | |
| Part (b) | ||
| Indicative content, one mark per point: | — | |
| at release only weight (down) and upthrust (up) act; weight is greater, so the bead accelerates downwards | ✓ 1 | |
| a viscous drag force acts upwards, opposing the motion, and it increases as the speed increases | ✓ 1 | |
| so the resultant force, weight − upthrust − drag, decreases and the acceleration decreases | ✓ 1 | |
| when weight = upthrust + drag, the resultant force is zero, so the acceleration is zero and the bead moves at terminal velocity | ✓ 1 | |
| warmer honey has a lower viscosity | ✓ 1 | |
| so the drag at a given speed is smaller and the bead must move faster before the forces balance: the terminal velocity is higher (since F = 6πηrv) | ✓ 1 | |
Exam tip: In an extended answer, link each step with “so” or “because”: speed rises → drag rises → resultant falls → acceleration falls → forces balance → constant velocity.
A student drops steel balls of several different radii r into a tall, wide tube of liquid soap and measures the terminal velocity v of each. She plots v (y-axis, in mm s−1) against r2 (x-axis, in mm2). The best-fit line passes through the origin and through the point (2.50 mm2, 61.5 mm s−1). The point for the largest ball lies clearly below the line. Density of steel = 7800 kg m−3; density of the liquid soap = 1030 kg m−3.
Explain why a graph of v against r2 is expected to be a straight line through the origin.
Determine the gradient of the line in SI units.
Calculate the viscosity of the liquid soap.
Suggest why the point for the largest ball lies below the line.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at terminal velocity weight = upthrust + drag, which gives v = [2g(ρs − ρf) ÷ 9η] × r2 | ✓ 1 | |
| the bracket is constant, so this has the form y = mx: v is proportional to r2 | ✓ 1 | |
| Part (b) | ||
| gradient = 61.5 ÷ 2.50 = 24.6 mm s−1 mm−2 | ✓ 1 | |
| = 24.6 × 10−3 ÷ 10−6 = 2.46 × 104 m−1 s−1 | ✓ 1 | |
| Part (c) | ||
| η = 2g(ρs − ρf) ÷ (9 × gradient) = 2 × 9.81 × 6770 ÷ (9 × 2.46 × 104) | ✓ 1 | |
| η = 0.600 Pa s | ✓ 1 | |
| Part (d) | ||
| any one: the flow round the large, fast ball is no longer laminar, so the drag is greater than Stokes' law predicts / the large ball had not yet reached terminal velocity when timing began / wall effects are larger for a large ball | ✓ 1 | |
Exam tip: Convert a gradient's units carefully: mm s−1 ÷ mm2 = 10−3 ÷ 10−6 = 103 m−1 s−1.
A sealed research balloon has an envelope that does not stretch, so its volume stays 3.0 m3. The envelope has a mass of 1.2 kg, the instruments 0.70 kg and the helium inside 0.50 kg; ignore the volume of the envelope and instruments. The graph shows how the density of air changes with height.
At about what height does the balloon stop rising?
| Answer | Mark | Notes |
|---|---|---|
| It stops rising where upthrust = weight: ρair × 3.0 × g = (1.2 + 0.70 + 0.50)g, so ρair = 2.4 ÷ 3.0 = 0.80 kg m−3. From the graph, this density occurs at 3.2 km. | ✓ 1 | Answer A |
Answer: A
Exam tip: Turn a force balance into a density: a body floats in a fluid where the fluid's density equals the body's total mass ÷ its volume. Then read that density off the graph.
The graph shows how the viscosity η of an oil depends on its temperature θ. A steel ball of radius 1.0 mm falls through the oil at 40 °C. Density of steel = 7800 kg m−3; density of the oil = 880 kg m−3. Stokes' law applies.
What is the terminal velocity of the ball?
| Answer | Mark | Notes |
|---|---|---|
| From the graph, η = 0.25 Pa s at 40 °C. Weight = upthrust + drag gives v = 2r2g(ρs − ρf) ÷ 9η = 2 × (1.0 × 10−3)2 × 9.81 × 6920 ÷ (9 × 0.25) = 0.0603 m s−1. | ✓ 1 | Answer D |
Answer: D
Exam tip: Derive the terminal velocity from W = U + F each time: (4/3)πr3(ρs − ρf)g = 6πηrv. The factor 2/9 comes from (4/3) ÷ 6.
The graph shows how the density of air at atmospheric pressure depends on its temperature. A hot-air balloon has an envelope of volume 2800 m3. On the day of a flight, the outside air is at 20 °C. The fabric of the envelope must not be heated above 120 °C. Ignore the volume of the basket, burner and passengers.
Use the graph to find the density of air at 20 °C and at 120 °C.
Explain why heating the air inside the envelope allows the balloon to rise.
Calculate the greatest total mass of envelope, basket, burner and passengers that the balloon can lift.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| at 20 °C: 1.20 kg m−3 (allow 1.19–1.21) | ✓ 1 | |
| at 120 °C: 0.90 kg m−3 (allow 0.89–0.91) | ✓ 1 | |
| Part (b) | ||
| the hot air is less dense, so the mass and weight of the air inside the envelope decrease | ✓ 1 | |
| the upthrust (the weight of cold air displaced) is unchanged, so once the total weight is less than the upthrust there is a resultant upward force | ✓ 1 | |
| Part (c) | ||
| upthrust = 1.20 × 2800 × 9.81 = 3.30 × 104 N | ✓ 1 | |
| weight of hot air at 120 °C = 0.90 × 2800 × 9.81 = 2.47 × 104 N | ✓ 1 | |
| mass = (3.30 − 2.47) × 104 ÷ 9.81 = 840 kg (allow 780–900 from readings) | ✓ 1 | |
Exam tip: The load a balloon can lift is (density of outside air − density of inside air) × volume. Read both densities carefully: a small reading error is a large fraction of the difference.
In a test rig, a sphere of radius 1.5 mm is pulled at different constant speeds v through a tank of liquid, and the drag force F on it is measured. The graph shows the results.
Use the straight part of the graph to determine the viscosity of the liquid.
Explain why the graph is not a straight line above about 0.25 m s−1.
A ball of the same radius, for which (weight − upthrust) = 12.0 mN, falls through the liquid. Use the graph to find its terminal velocity.
Calculate the terminal velocity that Stokes' law would predict for this ball, and explain why it differs from your answer to (c).
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| gradient = 6.0 × 10−3 N ÷ 0.20 m s−1 = 0.030 N s m−1 | ✓ 1 | |
| gradient = 6πηr | ✓ 1 | |
| η = 0.030 ÷ (6π × 1.5 × 10−3) = 1.1 Pa s (1.06) | ✓ 1 | |
| Part (b) | ||
| at higher speeds the flow round the sphere is no longer laminar / becomes turbulent | ✓ 1 | |
| Stokes' law (F ∝ v) only applies at low speeds in laminar flow, so the drag rises faster than in proportion to speed | ✓ 1 | |
| Part (c) | ||
| terminal velocity where F = 12.0 mN on the curve: 0.36 m s−1 (allow 0.35–0.37) | ✓ 1 | |
| Part (d) | ||
| v = 12.0 × 10−3 ÷ 0.030 = 0.40 m s−1 | ✓ 1 | |
| the real drag at these speeds is greater than Stokes' law gives, so the forces balance at a lower speed | ✓ 1 | |
Exam tip: Take the gradient from the straight section only, using points read at gridline crossings such as (0.20, 6.0). Convert mN to N before dividing.
The diagram shows apparatus set up to determine the viscosity of an oil by timing small steel balls falling through it. The positions of three rubber bands, X, Y and Z, can be read from the metre rule.
Describe how the student should use this apparatus, and any other equipment needed, to obtain an accurate value for the viscosity of the oil. Your answer should include how the measurements are used.
In one trial the ball takes 6.40 s to fall from Y to Z. Use the diagram to calculate the terminal velocity.
| Answer | Mark | Notes |
|---|---|---|
| Part (a) | ||
| Indicative content, one mark per point: | — | |
| measure the diameter of the ball with a micrometer, at several orientations, and halve the mean to find r | ✓ 1 | |
| find the densities of the steel and the oil, e.g. mass on a balance and volume from a measuring cylinder | ✓ 1 | |
| release the ball at the centre of the tube, just below the surface, so it falls away from the walls | ✓ 1 | |
| time the ball from X to Y and from Y to Z; terminal velocity is confirmed if the time for YZ is twice the time for XY, since YZ is twice XY | ✓ 1 | |
| repeat for several identical balls and average; keep the temperature constant, checking with the thermometer, because viscosity depends on temperature | ✓ 1 | |
| calculate v = distance ÷ time, then η = 2r2g(ρsteel − ρoil) ÷ 9v, from weight = upthrust + drag | ✓ 1 | |
| Part (b) | ||
| distance YZ = 90.0 − 50.0 = 40.0 cm = 0.400 m | ✓ 1 | |
| v = 0.400 ÷ 6.40 = 0.0625 m s−1 | ✓ 1 | |
Exam tip: In a method answer, say how each quantity is measured, with which instrument, and how it is used in the final equation. Name the check for terminal velocity explicitly.
We use essential cookies to make this site work. With your permission we'd also use analytics and advertising cookies to understand which resources help students most and how families find us. Nothing optional is set unless you accept. Privacy policy