Prerequisites
Before studying viscosity, make sure you understand:
- Fluid Pressure — Basics of fluid mechanics
- Newton’s Laws — Force and motion
What is Viscosity?
Viscosity is the property of a fluid that opposes the relative motion between its layers. It’s the internal friction in fluids.
Think of it as the “thickness” or “stickiness” of a fluid.
Pour honey and water from the same height:
- Water flows quickly (low viscosity)
- Honey flows slowly (high viscosity)
Honey has about 10,000 times the viscosity of water! On a cold day, honey becomes even more viscous — viscosity generally increases as temperature decreases.
Examples:
- Low viscosity: Water, air, alcohol
- Medium viscosity: Motor oil, blood
- High viscosity: Honey, glycerin, molten glass
Types of Fluid Flow
1. Streamline Flow (Laminar Flow)
Orderly flow where fluid particles move in parallel layers (laminas) with no mixing between layers.
Characteristics:
- Each particle follows a definite path (streamline)
- Velocity at any point remains constant with time
- No cross-currents or eddies
- Occurs at low velocities
Streamline: A curve whose tangent at any point gives the direction of fluid velocity at that point.
Properties of streamlines:
- Never cross each other
- Closer streamlines → higher velocity
- Farther streamlines → lower velocity
2. Turbulent Flow
Chaotic, irregular flow with eddies, vortices, and mixing between layers.
Characteristics:
- Velocity varies randomly with time
- Cross-currents and eddies present
- Occurs at high velocities
- Energy is dissipated as heat
Example: Smoke from a cigarette initially rises smoothly (laminar), then becomes chaotic (turbulent).
Reynolds Number
Reynolds number (Re) determines whether flow is laminar or turbulent:
$$\boxed{Re = \frac{\rho v D}{\eta}}$$where:
- ρ = density of fluid
- v = velocity of fluid
- D = characteristic dimension (diameter for pipes)
- η = coefficient of viscosity
Critical values:
- Re < 1000: Laminar flow
- Re > 2000: Turbulent flow
- 1000 < Re < 2000: Transition region
Interactive Demo: Visualize Fluid Flow
Watch how viscous fluids flow and transition from laminar to turbulent flow.
Reynolds number is dimensionless:
$$[Re] = \frac{[ML^{-3}][LT^{-1}][L]}{[ML^{-1}T^{-1}]} = \frac{[ML^{-1}T^{-1}]}{[ML^{-1}T^{-1}]} = 1$$It’s a pure ratio!
Coefficient of Viscosity (η)
When a fluid flows, layers move with different velocities. Viscous force opposes this relative motion.
Newton’s Law of Viscosity:
Consider two parallel layers of fluid separated by distance dy, moving with velocity difference dv.
Viscous force:
$$\boxed{F = -\eta A \frac{dv}{dy}}$$where:
- η = coefficient of viscosity
- A = area of contact between layers
- dv/dy = velocity gradient (rate of change of velocity)
Negative sign: Force opposes relative motion.
Coefficient of viscosity:
$$\boxed{\eta = \frac{F \cdot dy}{A \cdot dv}}$$SI Unit: Pa·s (Pascal-second) or N·s/m² or kg/(m·s)
CGS Unit: Poise (P)
- 1 Poise = 0.1 Pa·s
- 1 centipoise (cP) = 0.01 Poise = 10⁻³ Pa·s
Dimensions: [ML⁻¹T⁻¹]
Viscosity units: Think “Pa-se” → Pa·s (Pascal-second)
Water at 20°C: η ≈ 1 centipoise = 10⁻³ Pa·s
Typical Viscosity Values (at 20°C)
| Fluid | Viscosity (Pa·s) | Viscosity (cP) |
|---|---|---|
| Air | 1.8 × 10⁻⁵ | 0.018 |
| Water | 1.0 × 10⁻³ | 1.0 |
| Blood | 3-4 × 10⁻³ | 3-4 |
| Olive oil | 0.08 | 80 |
| Glycerin | 1.5 | 1500 |
| Honey | 10 | 10,000 |
| Motor oil (SAE 30) | 0.2-0.5 | 200-500 |
Effect of Temperature on Viscosity
Liquids: Viscosity decreases with increasing temperature
- Molecules move faster, overcome intermolecular forces more easily
- Example: Honey flows more easily when warm
Gases: Viscosity increases with increasing temperature
- More molecular collisions transfer momentum between layers
- Opposite behavior from liquids!
Motor oil is labeled SAE 10W-30, SAE 20W-50, etc.
- W stands for “Winter”
- First number (10W): viscosity at low temperature
- Second number (30): viscosity at high temperature
Multi-grade oils maintain relatively constant viscosity across temperature range — crucial for engine protection in cold starts and hot running conditions!
Stokes’ Law (Viscous Force on Sphere)
When a small sphere moves through a viscous fluid at low velocity, the viscous drag force is:
$$\boxed{F = 6\pi \eta r v}$$where:
- η = coefficient of viscosity
- r = radius of sphere
- v = velocity of sphere
Conditions for Stokes’ Law:
- Sphere must be small and perfectly spherical
- Velocity must be small (Re < 1, laminar flow)
- Fluid must be infinite in extent (or container much larger than sphere)
- No slip at surface (fluid layer at surface moves with sphere)
Derivation by Dimensional Analysis
Let F depend on η, r, v:
$$F = k \eta^a r^b v^c$$Dimensions:
$$[MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c$$ $$[MLT^{-2}] = [M^a L^{-a+b+c} T^{-a-c}]$$Equating powers:
- M: 1 = a → a = 1
- T: -2 = -a - c → -2 = -1 - c → c = 1
- L: 1 = -a + b + c → 1 = -1 + b + 1 → b = 1
Stokes showed experimentally and theoretically: k = 6π
$$\boxed{F = 6\pi \eta r v}$$A spherical raindrop of radius 1 mm falls through air (η = 1.8 × 10⁻⁵ Pa·s) at 4 m/s. Find viscous drag force.
Solution:
$$F = 6\pi \eta r v$$ $$= 6 \times 3.14 \times 1.8 \times 10^{-5} \times 10^{-3} \times 4$$ $$= 6 \times 3.14 \times 1.8 \times 4 \times 10^{-8}$$ $$= 1.36 \times 10^{-6} \text{ N}$$F ≈ 1.4 μN (micro-newtons)
Very small force, but significant for tiny droplets!
Terminal Velocity
When a body falls through a viscous fluid:
- Initially: Weight > Drag, body accelerates downward
- Gradually: Velocity increases → Drag increases (F ∝ v)
- Finally: Weight = Drag + Buoyancy, constant velocity achieved
This constant velocity is called terminal velocity.
Force balance at terminal velocity:
$$mg = F_B + F_{viscous}$$ $$\rho_{sphere} \cdot V \cdot g = \rho_{fluid} \cdot V \cdot g + 6\pi \eta r v_t$$ $$\rho_{sphere} \cdot \frac{4}{3}\pi r^3 \cdot g = \rho_{fluid} \cdot \frac{4}{3}\pi r^3 \cdot g + 6\pi \eta r v_t$$ $$\frac{4}{3}\pi r^3 g (\rho_{sphere} - \rho_{fluid}) = 6\pi \eta r v_t$$ $$\boxed{v_t = \frac{2r^2 g (\rho_{sphere} - \rho_{fluid})}{9\eta}}$$Simplified (when ρ_sphere » ρ_fluid, like lead ball in water):
$$\boxed{v_t = \frac{2r^2 \rho g}{9\eta}}$$Terminal velocity:
- ∝ r² (radius squared) — bigger drops fall much faster!
- ∝ (ρ_sphere - ρ_fluid) — denser objects fall faster
- ∝ 1/η — higher viscosity → slower fall
- ∝ g — terminal velocity would be different on Moon!
Large raindrop (r = 2 mm): v_t ≈ 9 m/s (falls fast, hits hard)
Drizzle (r = 0.2 mm): v_t ≈ 0.09 m/s (falls gently)
Since v_t ∝ r², a drop 10 times smaller falls 100 times slower!
This is why clouds float (tiny droplets with v_t < 0.01 m/s), but heavy rain falls quickly. Without viscosity, even tiny droplets would hit like bullets!
Time to Reach Terminal Velocity
For a sphere starting from rest:
$$\boxed{t = \frac{m}{6\pi \eta r} \ln\left(\frac{v_t}{v_t - v}\right)}$$Typically, terminal velocity is reached in fraction of a second for small objects.
Approximate time to reach 90% of terminal velocity:
$$t_{90\%} \approx \frac{2m}{6\pi \eta r} = \frac{m}{3\pi \eta r}$$A steel ball (ρ = 7800 kg/m³) of radius 1 mm falls through glycerin (η = 1.5 Pa·s, ρ = 1260 kg/m³). Find terminal velocity. (g = 10 m/s²)
Solution:
$$v_t = \frac{2r^2 g (\rho_{steel} - \rho_{glycerin})}{9\eta}$$ $$= \frac{2 \times (10^{-3})^2 \times 10 \times (7800 - 1260)}{9 \times 1.5}$$ $$= \frac{2 \times 10^{-6} \times 10 \times 6540}{13.5}$$ $$= \frac{0.1308}{13.5} = 0.0097 \text{ m/s}$$v_t ≈ 9.7 mm/s (very slow!)
Glycerin is very viscous, so even steel falls slowly.
Poiseuille’s Law (Flow Through Pipe)
For laminar flow of incompressible fluid through cylindrical pipe:
Volume flow rate:
$$\boxed{Q = \frac{\pi r^4 (P_1 - P_2)}{8\eta L}}$$where:
- Q = volume flow rate (m³/s)
- r = radius of pipe
- P₁ - P₂ = pressure difference between ends
- η = viscosity
- L = length of pipe
Key observations:
- Q ∝ r⁴ — Doubling radius increases flow by 16 times!
- Q ∝ ΔP — Linear dependence on pressure
- Q ∝ 1/η — Higher viscosity → lower flow
- Q ∝ 1/L — Longer pipe → lower flow
Blood flow in arteries follows Poiseuille’s law.
If an artery is 50% blocked (radius reduced to half):
- Flow rate reduces to (1/2)⁴ = 1/16 of normal!
This is why even partial artery blockage is dangerous — it drastically reduces blood flow. A mere 16% reduction in radius cuts flow by half!
This explains why doctors take arterial plaque so seriously.
Common Mistakes to Avoid
Wrong: Stokes’ law applies to all objects in fluids.
Correct: Stokes’ law applies only to:
- Small spherical objects
- Low velocities (laminar flow, Re < 1)
- Infinite medium (or very large container)
At high velocities, drag ∝ v² (not v), and F ≠ 6πηrv
Wrong: Terminal velocity is the maximum velocity possible.
Correct: Terminal velocity is the constant velocity when forces balance. An object initially moving faster than v_t will decelerate to v_t.
Wrong: v_t = 2r²ρg/(9η) always.
Correct: This is only when ρ_object » ρ_fluid. Generally:
$$v_t = \frac{2r^2 g (\rho_{object} - \rho_{fluid})}{9\eta}$$For less dense objects (like air bubbles in water), buoyancy is crucial!
Practice Problems
Level 1: JEE Main Basics
A sphere of radius 2 mm moves through water (η = 10⁻³ Pa·s) at 5 cm/s. Find the viscous force.
Solution:
$$F = 6\pi \eta r v$$ $$= 6 \times 3.14 \times 10^{-3} \times 2 \times 10^{-3} \times 5 \times 10^{-2}$$ $$= 6 \times 3.14 \times 10^{-8}$$ $$= 1.88 \times 10^{-7} \text{ N} = \boxed{0.188 \text{ μN}}$$Find the terminal velocity of a rain drop of radius 0.3 mm falling through air. (ρ_water = 1000 kg/m³, η_air = 1.8 × 10⁻⁵ Pa·s, neglect buoyancy, g = 10 m/s²)
Solution:
$$v_t = \frac{2r^2 \rho g}{9\eta}$$ $$= \frac{2 \times (3 \times 10^{-4})^2 \times 1000 \times 10}{9 \times 1.8 \times 10^{-5}}$$ $$= \frac{2 \times 9 \times 10^{-8} \times 10^4}{16.2 \times 10^{-5}}$$ $$= \frac{1.8 \times 10^{-3}}{16.2 \times 10^{-5}} = \frac{1.8}{0.162} = 11.1 \text{ m/s}$$v_t ≈ 11 m/s (about 40 km/h!)
Level 2: JEE Main/Advanced
Two spheres of same material and radii r and 2r fall through same viscous fluid. Find ratio of their terminal velocities.
Solution:
$$v_t = \frac{2r^2 \rho g}{9\eta}$$Terminal velocity ∝ r²
$$\frac{v_{t1}}{v_{t2}} = \frac{r_1^2}{r_2^2} = \frac{r^2}{(2r)^2} = \frac{1}{4}$$Ratio = 1:4
Larger sphere falls 4 times faster!
A liquid flows through a tube at 10 cm³/s when pressure difference is 20 Pa. What pressure difference is needed for flow rate of 25 cm³/s?
Solution:
From Poiseuille’s law: Q ∝ ΔP
$$\frac{Q_2}{Q_1} = \frac{\Delta P_2}{\Delta P_1}$$ $$\frac{25}{10} = \frac{\Delta P_2}{20}$$ $$\Delta P_2 = \frac{25 \times 20}{10} = \boxed{50 \text{ Pa}}$$Level 3: JEE Advanced
A sphere of density 2000 kg/m³ is dropped in a liquid of density 1500 kg/m³ and viscosity 1 Pa·s. If it attains terminal velocity of 1 cm/s, find its radius. (g = 10 m/s²)
Solution:
$$v_t = \frac{2r^2 g (\rho_{sphere} - \rho_{liquid})}{9\eta}$$ $$0.01 = \frac{2r^2 \times 10 \times (2000 - 1500)}{9 \times 1}$$ $$0.01 = \frac{2r^2 \times 10 \times 500}{9}$$ $$0.01 = \frac{10000 r^2}{9}$$ $$r^2 = \frac{0.01 \times 9}{10000} = \frac{0.09}{10000} = 9 \times 10^{-6}$$ $$r = 3 \times 10^{-3} \text{ m} = \boxed{3 \text{ mm}}$$Eight identical droplets of radius r and terminal velocity v coalesce to form a single drop. Find terminal velocity of larger drop.
Solution:
Volume conserved:
$$8 \times \frac{4}{3}\pi r^3 = \frac{4}{3}\pi R^3$$ $$R^3 = 8r^3$$ $$R = 2r$$Since v_t ∝ r²:
$$\frac{v_t(\text{large})}{v_t(\text{small})} = \frac{R^2}{r^2} = \frac{(2r)^2}{r^2} = 4$$v_t(large) = 4v
The combined drop falls 4 times faster!
Related Topics
Within Properties of Matter
- Fluid Pressure — Pressure in static fluids
- Bernoulli’s Theorem — Pressure in moving fluids
- Surface Tension — Surface effects
Connected Chapters
- Newton’s Laws — Force and drag
- Circular Motion — Centripetal force in viscous medium