Physics Fluids and Thermal - How to Solve: Viscosity & Terminal Velocity (Stokes’ Law) – IIT JEE Guide

How to Solve: Viscosity & Terminal Velocity (Stokes’ Law) – IIT JEE Guide

Introduction

Master Stokes’ Law, and you unlock 3-5 marks in IIT JEE—every year. A tiny steel ball dropped in oil? A raindrop falling from the sky? Even drug delivery in medicine—all use terminal velocity and Stokes’ Law. Miss this, and you lose easy marks on fluid dynamics questions.

WHAT YOU NEED TO KNOW FIRST

Before you start, make sure you understand:
1. Newton’s 2nd Law (F = ma) – Forces cause acceleration.
2. Free-body diagrams – How to draw forces on an object.
3. Viscosity (η) – A fluid’s internal friction (like honey vs. water).

If any of these are shaky, pause and review them first.

KEY TERMS & FORMULAS

Key Terms

1. Viscosity (η) – Resistance of a fluid to flow. Unit: Pa·s (Pascal-second) or poise (1 poise = 0.1 Pa·s).
2. Terminal Velocity (vₜ) – Constant speed when drag force = weight (no acceleration).
3. Stokes’ Law – Drag force on a small spherical object moving slowly in a viscous fluid.

Formulas

STEP-BY-STEP METHOD

Follow these 5 steps for every Stokes’ Law problem.

Step 1: Identify the Forces

• Weight (W) = Downward (always).
• Buoyant Force (F_b) = Upward (always).
• Drag Force (Fₛ) = Upward (opposes motion).

Draw a free-body diagram:

   ↑ Fₛ (Drag)
   ↑ F_b (Buoyant)
   ↓ W (Weight)

Step 2: Write the Force Balance at Terminal Velocity

At terminal velocity, net force = 0 (no acceleration). So: Fₛ + F_b = W

Step 3: Plug in the Formulas

• Fₛ = 6πηrvₜ (Stokes’ Law)
• F_b = (4/3)πr³σg (Buoyant Force)
• W = (4/3)πr³ρg (Weight)

Substitute into the force balance: 6πηrvₜ + (4/3)πr³σg = (4/3)πr³ρg

Step 4: Solve for Terminal Velocity (vₜ)

1. Subtract F_b from both sides: 6πηrvₜ = (4/3)πr³ρg – (4/3)πr³σg
2. Factor out (4/3)πr³g: 6πηrvₜ = (4/3)πr³g (ρ – σ)
3. Divide both sides by 6πηr: vₜ = [(4/3)πr³g (ρ – σ)] / (6πηr)
4. Simplify: vₜ = (2/9) × (ρ – σ)gr² / η

Step 5: Plug in Numbers & Calculate

• Check units (all in SI: kg, m, s, Pa·s).
• Calculate carefully (exam traps hide here!).

WORKED EXAMPLES

Example 1 – Basic (Direct Application)

Question: A steel ball (density = 7800 kg/m³, radius = 1 mm) falls in glycerine (viscosity = 1.5 Pa·s, density = 1260 kg/m³). Find its terminal velocity.

Solution: Step 1: Forces → W (down), F_b (up), Fₛ (up). Step 2: At terminal velocity: Fₛ + F_b = W. Step 3: Plug in formulas: 6πηrvₜ + (4/3)πr³σg = (4/3)πr³ρg Step 4: Solve for vₜ: vₜ = (2/9) × (ρ – σ)gr² / η ρ = 7800 kg/m³, σ = 1260 kg/m³, r = 1 mm = 10⁻³ m, η = 1.5 Pa·s, g = 9.8 m/s² vₜ = (2/9) × (7800 – 1260) × 9.8 × (10⁻³)² / 1.5 vₜ = (2/9) × 6540 × 9.8 × 10⁻⁶ / 1.5 vₜ = (2/9) × 0.064092 / 1.5 vₜ = 0.0095 m/s ≈ 9.5 mm/s

What we did and why: We used the force balance at terminal velocity and Stokes’ Law to find vₜ. The key was unit conversion (mm → m) and correct substitution.

Example 2 – Medium (Missing Density)

Question: A spherical oil drop (radius = 0.5 mm) falls in air (viscosity = 1.8 × 10⁻⁵ Pa·s, density = 1.2 kg/m³). Its terminal velocity is 0.1 m/s. Find the density of the oil drop.

Solution: Step 1: Forces → W (down), F_b (up), Fₛ (up). Step 2: At terminal velocity: Fₛ + F_b = W. Step 3: Plug in formulas: 6πηrvₜ + (4/3)πr³σg = (4/3)πr³ρg Step 4: Solve for ρ: 6πηrvₜ = (4/3)πr³g (ρ – σ) ρ = (6πηrvₜ × 3) / (4πr³g) + σ ρ = (18ηvₜ) / (4r²g) + σ η = 1.8 × 10⁻⁵ Pa·s, vₜ = 0.1 m/s, r = 0.5 mm = 5 × 10⁻⁴ m, σ = 1.2 kg/m³, g = 9.8 m/s² ρ = (18 × 1.8 × 10⁻⁵ × 0.1) / (4 × (5 × 10⁻⁴)² × 9.8) + 1.2 ρ = (3.24 × 10⁻⁵) / (9.8 × 10⁻⁶) + 1.2 ρ = 3.306 + 1.2 = 4.506 kg/m³

What we did and why: We rearranged the terminal velocity formula to solve for density (ρ). The trick was isolating ρ before plugging in numbers.

Example 3 – Exam-Style (Disguised Question)

Question (IIT JEE 2018 Adapted): A small metal sphere (density = 8000 kg/m³) is dropped in a viscous liquid (density = 1000 kg/m³, viscosity = 0.1 Pa·s). It reaches terminal velocity in 0.5 s. If the sphere’s radius is 2 mm, find the distance fallen before reaching terminal velocity. (Assume acceleration is constant until terminal velocity.)

Solution: Step 1: Find vₜ first. vₜ = (2/9) × (ρ – σ)gr² / η ρ = 8000 kg/m³, σ = 1000 kg/m³, r = 2 mm = 2 × 10⁻³ m, η = 0.1 Pa·s, g = 9.8 m/s² vₜ = (2/9) × (8000 – 1000) × 9.8 × (2 × 10⁻³)² / 0.1 vₜ = (2/9) × 7000 × 9.8 × 4 × 10⁻⁶ / 0.1 vₜ = (2/9) × 0.2744 / 0.1 = 0.61 m/s

Step 2: Use v = u + at to find acceleration. u = 0, v = vₜ = 0.61 m/s, t = 0.5 s a = (v – u) / t = 0.61 / 0.5 = 1.22 m/s²

Step 3: Find distance using s = ut + ½at². s = 0 + ½ × 1.22 × (0.5)² = 0.1525 m ≈ 15.3 cm

What we did and why: This was a two-part question—first Stokes’ Law, then kinematics. The trap was assuming terminal velocity instantly—we had to calculate acceleration first.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second Stokes’ Law survival guide."

1. Terminal velocity = when drag + buoyant force = weight.
2. Stokes’ Law: Fₛ = 6πηrv. Memorise it.
3. Terminal velocity formula: vₜ = (2/9)(ρ – σ)gr² / η. Memorise it.
4. Always draw a free-body diagram first.
5. Check units—convert mm to m, poise to Pa·s.
6. If they ask for distance before terminal velocity, use kinematics (v = u + at, s = ut + ½at²).
7. Watch for traps: non-spheres, missing densities, instant terminal velocity.

"You’ve got this. Now go crush that fluid dynamics question."