Specification: AQA A-level Physics (7408) 3.12.1.4 Turning points option | Paper: 3 Section B | Data and formulae booklet: F = 6πηrv is given in the exam (alongside QV/d = mg); the sphere volume (4/3)πr³ is with the geometrical equations; the terminal speed condition mg = 6πηrv is not, so learn it
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📌 KEY DEFINITIONS – learn them word for word
Stokes’ law: the viscous drag force on a small sphere of radius r moving at speed v through a fluid of viscosity η is F = 6πηrv. It applies to small spheres moving slowly, so the flow around them is streamlined (laminar).
Terminal speed: the constant speed reached when the drag force equals the weight (ignoring upthrust), so the resultant force is zero: mg = 6πηrv.
Finding r in Millikan’s experiment: with the plates switched off, the drop falls at its terminal speed. Using m = (4/3)πr³ρ gives r = √(9ηv ÷ 2ρg), and then m. The charge follows from QV/d = mg when the drop is held stationary.
The equation
viscous force = 6π × viscosity × radius × speed F = 6πηrv at terminal speed, with m = (4/3)πr³ρ: (4/3)πr³ρg = 6πηrv
| Quantity | Symbol | Unit |
|---|---|---|
| viscous (drag) force | F | newtons (N) |
| viscosity of the fluid | η | N s m⁻² (equivalent to Pa s) |
| radius of the sphere | r | metres (m) |
| speed of the sphere | v | metres per second (m s⁻¹) |
| density of the drop | ρ | kilograms per cubic metre (kg m⁻³) |
Rearranged: r = F ÷ 6πηv | η = F ÷ 6πrv | terminal speed v = 2ρgr² ÷ 9η | r = √(9ηv ÷ 2ρg) | Convert μm to m (× 10−6) and mm s⁻¹ to m s⁻¹ (× 10−3). Ignore upthrust unless the question gives you the density of the fluid.
Use FIFA for every answer
- F – Formula: write the equation as it appears on the sheet.
- I – Insert: put in the numbers (convert to standard units first).
- F – Fix: rearrange to make the unknown the subject.
- A – Answer: calculate and always give the unit.
Core: Q1–4. Exam standard: Q5–8. Stretch: Q9–10.
Questions
Q1 (Core) An oil drop of radius 1.5 × 10−6 m falls through air at 2.4 × 10−4 m s⁻¹. Calculate the viscous force on the drop, to 2 significant figures. (ηair = 1.8 × 10−5 N s m⁻²)
Show solution
F: F = 6πηrv
I: F = 6π × 1.8 × 10−5 × 1.5 × 10−6 × 2.4 × 10−4
F: F is already the subject
A: 1.22… × 10−13 N = 1.2 × 10−13 N (2 s.f.)
Q2 (Core) A small sphere of radius 1.0 × 10−3 m moves at 0.020 m s⁻¹ through glycerol of viscosity 1.4 N s m⁻². Calculate the viscous force on it, to 2 significant figures.
Show solution
F: F = 6πηrv
I: F = 6π × 1.4 × 1.0 × 10−3 × 0.020
F: F is already the subject
A: 5.27… × 10−4 N = 5.3 × 10−4 N (2 s.f.)
Q3 (Core) An oil drop falls through air at a terminal speed of 0.18 mm s⁻¹. The viscous force on it is 7.9 × 10−14 N. Calculate the radius of the drop in μm, to 2 significant figures. (ηair = 1.8 × 10−5 N s m⁻²)
Show solution
Convert first: 0.18 mm s⁻¹ = 1.8 × 10−4 m s⁻¹
F: F = 6πηrv
I: 7.9 × 10−14 = 6π × 1.8 × 10−5 × r × 1.8 × 10−4
F: r = F ÷ 6πηv = 7.9 × 10−14 ÷ 6.107… × 10−8
A: 1.29… × 10−6 m = 1.3 μm (2 s.f.)
Watch out: Stokes’ law uses the radius. If a question gives the diameter, halve it first.
Q4 (Core) A small sphere of radius 2.0 mm falls through a liquid. At a speed of 3.5 cm s⁻¹ the viscous force on it is 1.85 × 10−3 N. Calculate the viscosity of the liquid, to 3 significant figures.
Show solution
Convert first: 2.0 mm = 2.0 × 10−3 m; 3.5 cm s⁻¹ = 0.035 m s⁻¹
F: F = 6πηrv
I: 1.85 × 10−3 = 6π × η × 2.0 × 10−3 × 0.035
F: η = F ÷ 6πrv = 1.85 × 10−3 ÷ 1.319… × 10−3
A: 1.40 N s m⁻² (1.40 Pa s, 3 s.f.)
Q5 (Exam) (a) Show that the terminal speed of an oil drop falling through air is v = 2ρgr² ÷ 9η. (b) Calculate the terminal speed of a drop of radius 1.0 μm, to 2 significant figures. (c) Calculate how long the drop takes to fall 1.0 mm at this speed, to 2 significant figures. Ignore upthrust. (ηair = 1.8 × 10−5 N s m⁻², ρoil = 880 kg m⁻³, g = 9.81 m s⁻²)
Show solution
Step 1 – (a) derivation
at terminal speed, weight = drag: mg = 6πηrv with m = (4/3)πr³ρ, so (4/3)πr³ρg = 6πηrv and v = 2ρgr² ÷ 9η
Convert first: 1.0 μm = 1.0 × 10−6 m; 1.0 mm = 1.0 × 10−3 m
Step 2 – (b) terminal speed
I: v = (2 × 880 × 9.81 × (1.0 × 10−6)²) ÷ (9 × 1.8 × 10−5)
A: 1.06… × 10−4 m s⁻¹ = 1.1 × 10−4 m s⁻¹ (2 s.f.)
Step 3 – (c) time to fall 1.0 mm
t = 1.0 × 10−3 ÷ 1.06… × 10−4 = 9.38… s
A: 9.4 s (2 s.f.), slow enough to time with a stopwatch through the microscope
Watch out: r is squared in v = 2ρgr² ÷ 9η but appears only once in F = 6πηrv. Cancel one r on each side carefully when you derive it.
Q6 (Exam) With the plates switched off, an oil drop in Millikan’s apparatus falls 0.50 mm at constant speed in 4.6 s. Calculate (a) its terminal speed, (b) its radius, to 2 significant figures, and (c) its mass, to 2 significant figures. Ignore upthrust. (ηair = 1.8 × 10−5 N s m⁻², ρoil = 880 kg m⁻³, g = 9.81 m s⁻²)
Show solution
Convert first: 0.50 mm = 5.0 × 10−4 m
Step 1 – (a) terminal speed
v = 5.0 × 10−4 ÷ 4.6 = 1.086… × 10−4 m s⁻¹
Step 2 – (b) radius
F: (4/3)πr³ρg = 6πηrv, so r = √(9ηv ÷ 2ρg)
I: r = √((9 × 1.8 × 10−5 × 1.086… × 10−4) ÷ (2 × 880 × 9.81))
A: 1.009… × 10−6 m = 1.0 μm (2 s.f.)
Step 3 – (c) mass
m = (4/3)πr³ρ = (4/3)π × (1.009… × 10−6)³ × 880
A: 3.79… × 10−15 kg = 3.8 × 10−15 kg (2 s.f.)
Q7 (Exam) Two drops of the same oil fall through air with the plates switched off. Drop X has a terminal speed of 2.0 × 10−4 m s⁻¹ and drop Y has a terminal speed of 5.0 × 10−5 m s⁻¹. Calculate (a) the ratio of their radii, rX ÷ rY, and (b) the ratio of their masses, mX ÷ mY. Ignore upthrust.
Show solution
Step 1 – how v depends on r
(4/3)πr³ρg = 6πηrv gives v = 2ρgr² ÷ 9η, so v ∝ r² for the same oil in the same air
Step 2 – (a) radii
vX ÷ vY = 2.0 × 10−4 ÷ 5.0 × 10−5 = 4, so rX ÷ rY = √4
A: 2
Step 3 – (b) masses
m ∝ r³, so mX ÷ mY = 2³
A: 8
Watch out: v is not proportional to r. A drop that falls four times as fast is only twice as wide, but eight times as heavy.
Q8 (Exam) Calculate the terminal speed of an oil drop of radius 1.5 μm falling through air, to 3 significant figures, (a) ignoring upthrust and (b) including the upthrust of the air, density 1.2 kg m⁻³. (c) Calculate the percentage difference, to 2 significant figures, and comment. (ηair = 1.8 × 10−5 N s m⁻², ρoil = 880 kg m⁻³, g = 9.81 m s⁻²)
Show solution
Convert first: 1.5 μm = 1.5 × 10−6 m
Step 1 – (a) ignoring upthrust
F: mg = 6πηrv, so v = 2ρgr² ÷ 9η
I: v = (2 × 880 × 9.81 × (1.5 × 10−6)²) ÷ (9 × 1.8 × 10−5)
A: 2.398 × 10−4 m s⁻¹ = 2.40 × 10−4 m s⁻¹ (3 s.f.)
Step 2 – (b) including upthrust
weight − upthrust = drag: (4/3)πr³(ρoil − ρair)g = 6πηrv, so replace 880 by 880 − 1.2 = 878.8 kg m⁻³
v = 2.398 × 10−4 × 878.8 ÷ 880 = 2.394… × 10−4 m s⁻¹
A: 2.39 × 10−4 m s⁻¹ (3 s.f.)
Step 3 – (c) comparison
percentage difference = 1.2 ÷ 880 × 100 = 0.14% (2 s.f.). This is tiny because air is about 700 times less dense than oil, which is why upthrust is usually ignored.
Q9 (Stretch) In Millikan’s experiment, with the plates switched off, a drop falls at a terminal speed of 1.20 × 10−4 m s⁻¹. With a pd of 450 V across plates 5.0 mm apart, the same drop is held stationary. Calculate (a) the radius of the drop, (b) its mass, (c) its charge, each to 3 significant figures, and (d) the number of electron charges it carries. Ignore upthrust. (ηair = 1.8 × 10−5 N s m⁻², ρoil = 880 kg m⁻³, g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Convert first: 5.0 mm = 5.0 × 10−3 m
Step 1 – (a) radius from the terminal speed (plates off)
F: (4/3)πr³ρg = 6πηrv, so r = √(9ηv ÷ 2ρg)
I: r = √((9 × 1.8 × 10−5 × 1.20 × 10−4) ÷ (2 × 880 × 9.81)) = √(1.125… × 10−12)
A: 1.061… × 10−6 m = 1.06 μm (3 s.f.)
Step 2 – (b) mass
m = (4/3)πr³ρ = (4/3)π × (1.061… × 10−6)³ × 880 = 4.403… × 10−15 kg
A: 4.40 × 10−15 kg (3 s.f.)
Step 3 – (c) charge from the stationary drop (plates on)
F: QV/d = mg, so Q = mgd ÷ V
I: Q = (4.403… × 10−15 × 9.81 × 5.0 × 10−3) ÷ 450
A: 4.800… × 10−19 C = 4.80 × 10−19 C (3 s.f.)
Step 4 – (d) number of charges
n = 4.800… × 10−19 ÷ 1.60 × 10−19 = 3.00…, so 3 electron charges
Watch out: when the drop is stationary there is no viscous force, so Stokes’ law is only used for the falling drop with the plates off.
Q10 (Stretch) With the plates switched off, an oil drop falls at a terminal speed v₁ = 9.0 × 10−5 m s⁻¹. A pd of 1000 V is then applied across plates 6.0 mm apart, and the drop rises at a constant speed v₂ = 8.1 × 10−5 m s⁻¹. (a) Show that QV/d = 6πηr(v₁ + v₂). (b) Calculate the radius of the drop, to 3 significant figures. (c) Calculate the charge on the drop, to 3 significant figures, and deduce the number of electron charges. Ignore upthrust. (ηair = 1.8 × 10−5 N s m⁻², ρoil = 880 kg m⁻³, g = 9.81 m s⁻², e = 1.60 × 10−19 C)
Show solution
Step 1 – (a) the force equations
falling, plates off: mg = 6πηrv₁. Rising at constant speed, the drag acts downwards with the weight: QV/d = mg + 6πηrv₂
substitute for mg: QV/d = 6πηrv₁ + 6πηrv₂ = 6πηr(v₁ + v₂)
Convert first: 6.0 mm = 6.0 × 10−3 m
Step 2 – (b) radius from v₁
r = √(9ηv₁ ÷ 2ρg) = √((9 × 1.8 × 10−5 × 9.0 × 10−5) ÷ (2 × 880 × 9.81))
A: 9.189… × 10−7 m = 0.919 μm (3 s.f.)
Step 3 – (c) charge
F: QV/d = 6πηr(v₁ + v₂)
I: Q × 1000 ÷ 6.0 × 10−3 = 6π × 1.8 × 10−5 × 9.189… × 10−7 × (9.0 × 10−5 + 8.1 × 10−5)
F: Q = 6πηr(v₁ + v₂) × d ÷ V
A: 3.19… × 10−19 C = 3.20 × 10−19 C (3 s.f.); n = 3.19… ÷ 1.60 = 1.99…, so 2 electron charges
Watch out: when the drop rises, the drag force points down. Adding it on the wrong side gives QV/d = 6πηr(v₁ − v₂) and a charge about 19 times too small.
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