Stokes' Law & Terminal VelocityJEE Main

Concept

A small sphere in a viscous fluid feels Stokes drag $F = 6\pi\eta r v$, growing with speed. It accelerates until drag + buoyancy balance weight — then falls at constant terminal velocity, which scales as $r^2$: big raindrops fall much faster than mist.

Key formula

Derivation

At terminal velocity the net force vanishes:

$$\underbrace{\tfrac{4}{3}\pi r^3\rho_s g}_{weight} = \underbrace{\tfrac{4}{3}\pi r^3\rho_f g}_{buoyancy} + \underbrace{6\pi\eta r v_t}_{drag}$$

Solve for $v_t$. The approach is exponential with time constant $\tau = m/6\pi\eta r$ — heavier spheres take longer to settle at $v_t$.

Scenarios to explore

• Stokes' Law — Terminal velocity of a sphere in a viscous fluid — v ∝ r².

Real-world applications

• Millikan's oil-drop experiment measured e using terminal velocity.
• Falling-ball viscometers measure η of oils.
• Sedimentation and centrifuge separation rates.

JEE exam tips

• v_t ∝ r²: doubling the radius quadruples terminal speed.
• Denser fluid or higher η ⇒ slower fall; if ρ_f > ρ_s the sphere RISES at the same formula's speed.
• Two drops coalescing: volume adds ⇒ r → 2^{1/3}r ⇒ v_t → 2^{2/3}·v_t.

Common mistakes

• Forgetting buoyancy — the density DIFFERENCE drives the fall.
• Applying Stokes' law at high speeds (it needs laminar flow, low Reynolds number).
• Mixing radius and diameter in the r² dependence.

Exam traps to avoid

• Drag depends on v, weight does not — only at v_t are they in balance.
• Raindrops hit at ~9 m/s, not the 500+ m/s free-fall would give from cloud height — viscosity saves us.