Viscosity, Stokes' Law and Surface Tension
Mechanical Properties of Fluids: Viscosity, Stokes' Law and Surface Tension
Viscosity, Stokes' Law and Surface Tension
Viscosity, Stokes' Law and Surface Tension
What you'll learn
- Define viscosity and apply Newton's law of viscous flow
- Use Stokes' law to find drag force on a sphere
- Derive terminal velocity and solve numerical problems
- Interpret the Reynolds number for laminar vs turbulent flow
- Define surface tension and apply formulas for bubbles, drops, and capillary rise
- Connect these concepts to real engineering and biological systems
Key concepts
Level 1 — Foundations
Viscosity
Viscosity η is the internal friction in a fluid that resists relative motion between layers.
Newton's Law of Viscosity:
where:
- F = tangential (shear) force (N)
- A = area of the layer (m²)
- dv/dy = velocity gradient perpendicular to flow (s⁻¹)
- η = coefficient of viscosity (Pa·s or N·s/m² or Poise)
SI unit: Pa·s. 1 Poise (CGS) = 0.1 Pa·s.
Honey has η ≈ 10 Pa·s; water at 20°C has η ≈ 0.001 Pa·s = 1 mPa·s.
Viscosity decreases with temperature in liquids (thermal agitation weakens intermolecular bonds). In gases, viscosity increases with temperature.
Stokes' Law
Drag force on a sphere of radius r moving at velocity v through a fluid of viscosity η:
Terminal Velocity
A sphere falling through a viscous fluid accelerates until drag + buoyancy = weight (net force = 0):
where ρ = sphere density, σ = fluid density.
- v_t ∝ r² (larger sphere falls much faster)
- v_t ∝ (ρ − σ) (denser sphere falls faster; sphere floats if ρ < σ)
- v_t ∝ 1/η (higher viscosity → lower terminal velocity)
Surface Tension
Surface tension T = force per unit length acting along the liquid surface tangentially:
SI unit: N/m. Arises because surface molecules have net inward intermolecular force — liquid minimises surface area.
Level 2 — JEE Depth
Terminal Velocity Derivation
At terminal velocity, net force = 0:
Weight (down) = Buoyancy (up) + Stokes drag (up)
For ρ > σ: sphere falls (v_t positive downward) For ρ < σ: sphere rises (v_t negative, i.e. upward — bubble rising in liquid)
Reynolds Number
Dimensionless parameter predicting flow regime:
where d = characteristic length (pipe diameter for pipe flow, sphere diameter for Stokes flow).
| Re | Flow type |
|---|---|
| < 1 | Stokes (creeping) flow — Stokes law valid |
| < 1000 | Laminar flow |
| 1000–2000 | Transitional |
| > 2000 | Turbulent flow |
JEE note: Stokes law is only valid for Re << 1. For large spheres/high speeds, drag ≠ 6πηrv.
Excess Pressure — Bubbles and Drops
Surface tension creates excess pressure inside curved surfaces:
| Surface | Excess pressure |
|---|---|
| Liquid drop (one surface) | ΔP = 2T/r |
| Soap bubble (two surfaces — inner + outer) | ΔP = 4T/r |
| Air bubble in liquid (one surface) | ΔP = 2T/r |
A soap bubble has two surfaces (inside wall and outside wall), hence the factor of 4 not 2.
Capillary Rise
When a liquid wets a narrow tube (contact angle θ < 90°), it rises to height h:
Force balance: upward surface tension force = weight of liquid column
- Wetting liquid (water on glass, θ ≈ 0°): rises (h positive)
- Non-wetting liquid (mercury on glass, θ ≈ 140°): depressed (h negative)
- h ∝ 1/r (finer tube → greater rise) — explains tree water transport in xylem
Jurin's Law: h × r = constant for a given liquid-solid-gas system.
Surface Energy
Work done to increase surface area by ΔA: W = T × ΔA Energy stored per unit area = T (same as surface tension numerically)
When two drops merge: energy released = T × (4πr₁² + 4πr₂² − 4πR²) where R is radius of merged drop (conserve volume: R³ = r₁³ + r₂³).
Worked example
Example 1: Sphere radius 1 mm, density 2500 kg/m³ falling in oil (η = 0.8 Pa·s, ρ_oil = 900 kg/m³), g = 10 m/s²
Given: r = 1 mm = 10⁻³ m, ρ = 2500 kg/m³, σ = 900 kg/m³
η = 0.8 Pa·s, g = 10 m/s²
Formula: v_t = 2r²(ρ − σ)g / (9η)
Step 1: Calculate numerator
2 × (10⁻³)² × (2500 − 900) × 10
= 2 × 10⁻⁶ × 1600 × 10
= 2 × 10⁻⁶ × 16000
= 32 × 10⁻³ = 0.032
Step 2: Calculate denominator
9 × 0.8 = 7.2
Step 3: Terminal velocity
v_t = 0.032 / 7.2 = 0.00444 m/s ≈ 4.4 mm/s
This is very slow — consistent with a small sphere in a thick oil.
A larger sphere (r = 2 mm) would fall at 4× this speed = 17.8 mm/s (v_t ∝ r²).
Example 2: Glass capillary radius 0.1 mm in water (T = 0.073 N/m, θ = 0°, ρ = 1000 kg/m³, g = 10 m/s²)
Given: r = 0.1 mm = 10⁻⁴ m, T = 0.073 N/m, θ = 0° → cos θ = 1
ρ = 1000 kg/m³, g = 10 m/s²
Formula: h = 2T cosθ / (ρgr)
h = (2 × 0.073 × 1) / (1000 × 10 × 10⁻⁴)
h = 0.146 / (1000 × 10 × 10⁻⁴)
h = 0.146 / 1
h = 0.146 m = 14.6 cm
Check units: [N/m] / ([kg/m³][m/s²][m]) = [N/m] / [N/m²] = m ✓
For a finer capillary of r = 0.05 mm: h doubles = 29.2 cm (h ∝ 1/r).
This is how trees draw water up from roots to leaves through fine xylem channels.
Common mistakes
| Mistake | Why it happens | Fix |
|---|---|---|
| Using ΔP = 2T/r for soap bubble instead of 4T/r | Forgetting soap bubble has two surfaces | Soap bubble: 4T/r (2 surfaces); liquid drop or air bubble in liquid: 2T/r (1 surface) |
| Forgetting buoyancy term in terminal velocity | Only thinking of drag vs gravity | Three forces act: weight down, buoyancy up, Stokes drag up; all three must be balanced |
| Using r in mm instead of metres in v_t formula | Unit carelessness | v_t ∝ r² — error of r in mm gives answer off by 10⁶; always convert to metres first |
| Thinking capillary rise increases with wider tube | Intuition that bigger tube holds more | h = 2Tcosθ/(ρgr): wider tube (larger r) gives smaller h — narrower is higher |
Quick check
- Q1 A sphere of radius 2 mm and density 3000 kg/m³ falls in a liquid (η = 1.0 Pa·s, ρ_liquid = 800 kg/m³). Find terminal velocity.
- Q2 Find the excess pressure inside (a) a water drop of radius 2 mm (T = 0.073 N/m), and (b) a soap bubble of the same radius (T = 0.04 N/m).
- Q3 A capillary of radius 0.5 mm holds water to height h. If a capillary of radius 0.25 mm is used, what is the new height?
- Q4 Calculate the Reynolds number for flow in a pipe of diameter 2 cm at 0.5 m/s with ρ = 1000 kg/m³, η = 0.001 Pa·s. Is the flow laminar or turbulent?
- Stretch: Q5 Two soap bubbles of radii 3 cm and 4 cm coalesce to form a single bubble at constant temperature. Find the radius of the new bubble. (Hint: conserve volume, not area.)
NCERT Chapter 9 link: Section 9.5 covers viscosity, Stokes law, and terminal velocity. Section 9.6 covers Reynolds number. Section 9.7 covers surface tension, surface energy, excess pressure, and capillary rise. All formulas and derivations are in NCERT — read them with pencil in hand, re-deriving each step.
Exam connections: JEE Main: terminal velocity calculation, excess pressure in bubbles/drops, capillary rise formula. JEE Advanced: comparing terminal velocities of spheres with different r and ρ, energy released when bubbles merge, capillary rise in tilted tubes, Reynolds number interpretation. The v_t ∝ r² dependency is frequently tested via ratio problems.
Study strategy: Write all formulas in a single table with units. For terminal velocity, practise recognising all three forces (weight, buoyancy, Stokes drag) from the free-body diagram — don't memorise formula mechanically. For surface tension, distinguish carefully between "one surface" (drop, air bubble) and "two surfaces" (soap bubble) before writing ΔP.
Interactive Exploration Suggestions (Drishti Live Worlds)
- Use the platform-native live simulation: a viscosity sandbox where you drop spheres of different radii and densities into fluids of different viscosities and observe terminal velocity values.
- Mirror / body / home activity: drop small balls of modelling clay (different sizes) into a jar of honey and measure how long they take to fall — verify v_t ∝ r² by comparing timings.
- Voice or text reflection with AI Mentor: explain to a younger sibling why a soap bubble is round and why a bigger bubble has lower excess pressure inside.
AI Mentor Prompts (Socratic, Board-Adaptive)
- "Explain why a small dust particle falls much more slowly than a marble, even though both experience the same gravitational acceleration — use the terminal velocity formula."
- "What is the difference between excess pressure inside a soap bubble and a water drop of the same radius? Why?"
- Stretch: "How do trees pull water from the soil up to leaves 10–20 m above, without any pump? Which physics principle makes this possible and what limits the height?"
Gamification, Portfolio & Parent Visibility
- Complete the core practice + one extension activity (photo, table, short reflection, or mini-project) for base XP + topic badge.
- 5-7 day streak or family discussion note = multiplier + visible artifact in parent/principal dashboard.
- Best real-world application stories (anonymised) featured on class or national leaderboard.
Robotics, STEM & Future Skills Bridges
- Design a viscometer: drop steel balls of known radius into a tall cylinder of oil, measure fall time over a fixed distance, and back-calculate η from v_t = distance/time.
- Future Skill track: Sustainable Living / Green Tech — viscosity and surface tension determine lubricant selection in engines, paint behaviour, and microfluidic chip design for diagnostics.
- Coding extension: Write a simulation that animates a sphere falling through a viscous fluid, updating velocity each time step using F_net = mg − buoyancy − Stokes drag, until v_t is reached. Plot v vs time.
NEP 2020 & Full Education OS Alignment
This material emphasises experiential "learning by doing", competency (apply/create/analyse), vocational exposure, critical thinking, and multidisciplinary connections. Designed to feed live worlds, AI Mentor (with memory), gamification, robotics, parent analytics, and future skills — not just exam prep.
Portfolio Evidence Idea: Your photo/table/reflection/project + one sentence on "How this helps me in real life or a possible future path."
Open the Practice tab for aligned questions (easy/medium/hard + case-based) with full AI scaffolding.
See curriculum for cross-links and the full future-skills/robotics chapters.
Key Takeaways (TL;DR)
- What you'll learn
- Key concepts
- Worked example
- Common mistakes
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