Viscosity — Stokes law and terminal velocity of a sphere falling in fluid

Question

A steel ball of radius 2 mm and density 8000 kg/m³ falls through glycerine of density 1300 kg/m³ and viscosity 0.83 Pa·s. Find the terminal velocity using Stokes' law. Also derive the expression for terminal velocity.

(JEE Main 2022, similar pattern)

Solution — Step by Step

Forces on the sphere

Three forces act on a sphere falling through a viscous fluid:

Weight (downward): $W = \frac{4}{3}\pi r^3 \rho_s g$
Buoyancy (upward): $F_b = \frac{4}{3}\pi r^3 \rho_f g$
Viscous drag (upward): $F_v = 6\pi \eta r v$ (Stokes' law)

where $\rho_s$ = density of sphere, $\rho_f$ = density of fluid, $\eta$ = viscosity.

Derive terminal velocity

At terminal velocity, acceleration = 0. So net force = 0:

$W = F_b + F_v$

$\frac{4}{3}\pi r^3 \rho_s g = \frac{4}{3}\pi r^3 \rho_f g + 6\pi \eta r v_t$

$6\pi \eta r v_t = \frac{4}{3}\pi r^3 (\rho_s - \rho_f) g$

Terminal Velocity

$v_t = \frac{2r^2(\rho_s - \rho_f)g}{9\eta}$

Substitute numerical values

$r = 2 \times 10^{-3}$ m, $\rho_s = 8000$ kg/m³, $\rho_f = 1300$ kg/m³, $\eta = 0.83$ Pa·s, $g = 9.8$ m/s².

$v_t = \frac{2 \times (2 \times 10^{-3})^2 \times (8000 - 1300) \times 9.8}{9 \times 0.83}$

$= \frac{2 \times 4 \times 10^{-6} \times 6700 \times 9.8}{7.47}$

$= \frac{2 \times 4 \times 6700 \times 9.8 \times 10^{-6}}{7.47} = \frac{525280 \times 10^{-6}}{7.47}$

$v_t \approx \mathbf{0.0703 \text{ m/s}} \approx 7.03 \text{ cm/s}$

Why This Works

When a sphere enters a viscous fluid, it initially accelerates due to gravity. But the viscous drag increases with velocity (Stokes' law: $F \propto v$ ). Eventually, the drag + buoyancy exactly balances the weight, and the sphere stops accelerating — it moves at constant "terminal" velocity.

Stokes' law ( $F = 6\pi\eta rv$ ) applies only for laminar flow (low Reynolds number, Re < 1). For larger speeds or larger objects, the flow becomes turbulent and the drag force is proportional to $v^2$ instead.

Key proportionality: $v_t \propto r^2$ . Doubling the radius increases terminal velocity 4 times. This explains why large raindrops fall faster than small ones, and why fine dust particles stay suspended for hours.

Alternative Method

You can use dimensional analysis to verify the formula: $[v_t] = [r^2][\rho][g]/[\eta] = \text{m}^2 \times \text{kg/m}^3 \times \text{m/s}^2 / (\text{Pa·s}) = \text{m/s}$ . The numerical coefficient (2/9) must come from the actual derivation.

💡 Expert Tip

For NEET, remember these proportionalities: $v_t \propto r^2$ , $v_t \propto (\rho_s - \rho_f)$ , $v_t \propto 1/\eta$ . MCQs often ask "if radius is doubled, what happens to terminal velocity?" Answer: it becomes 4 times. No calculation needed.

Common Mistake

⚠️ Common Mistake

Students often use $\rho_s$ instead of $(\rho_s - \rho_f)$ in the formula, forgetting the buoyancy correction. Without buoyancy, the terminal velocity is overestimated. The buoyant force is significant — in this problem, $\rho_f = 1300$ kg/m³ is about 16% of $\rho_s$ . Always subtract the fluid density from the sphere density.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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