Viscosity — Newton's law, Stokes' law, Reynolds number, streamline vs turbulent

Question

Define viscosity and state Newton’s law of viscous flow. A steel ball of radius 2 mm and density 7800 kg/m $^3$ falls through glycerine (density 1260 kg/m $^3$ , viscosity 0.83 Pa-s). Find the terminal velocity using Stokes’ law. What is the Reynolds number, and is the flow streamline or turbulent?

(JEE Main + NEET pattern)

Solution — Step by Step

The viscous force between two fluid layers is:

F = -\eta A \frac{dv}{dy}

where $\eta$ is the coefficient of viscosity (Pa-s or N-s/m $^2$ ), $A$ is the area of contact, and $\frac{dv}{dy}$ is the velocity gradient perpendicular to the flow.

Viscosity is the fluid’s internal resistance to flow. Honey has high viscosity; water has low viscosity.

When the ball reaches terminal velocity, the net force is zero:

Weight $=$ Buoyancy $+$ Viscous drag

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

Solving for terminal velocity:

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

v_t = \frac{2 \times (2 \times 10^{-3})^2 \times (7800 - 1260) \times 9.8}{9 \times 0.83}

v_t = \frac{2 \times 4 \times 10^{-6} \times 6540 \times 9.8}{7.47} = \frac{0.5127}{7.47}

v_t \approx \mathbf{0.069 \text{ m/s}} \approx 6.9 \text{ cm/s}

Re = \frac{\rho_f v d}{\eta} = \frac{1260 \times 0.069 \times 4 \times 10^{-3}}{0.83}

Re = \frac{0.348}{0.83} \approx \mathbf{0.42}

Since $Re \ll 1000$ (and even $\ll 1$ ), the flow is streamline (laminar). Stokes’ law is valid (it assumes $Re < 1$ ).

Rule of thumb: $Re < 2000$ indicates laminar flow; $Re > 4000$ indicates turbulent flow; between 2000-4000 is the transition zone.

flowchart TD
    A["Object falling through fluid"] --> B["Forces: weight, buoyancy, viscous drag"]
    B --> C["At terminal velocity: net force = 0"]
    C --> D["vt = 2r²(ρs - ρf)g / 9η"]
    D --> E["Calculate Reynolds number"]
    E --> F{"Re value?"}
    F -->|"Re < 2000"| G["Laminar / streamline flow"]
    F -->|"2000 < Re < 4000"| H["Transition zone"]
    F -->|"Re > 4000"| I["Turbulent flow"]
    G --> J["Stokes' law valid"]

Why This Works

Terminal velocity is reached when the accelerating force (gravity minus buoyancy) is exactly balanced by the retarding viscous force. At this point, acceleration is zero and the ball falls at constant speed.

Stokes’ law ( $F = 6\pi\eta rv$ ) applies specifically to a smooth sphere moving slowly through a viscous fluid. The assumption is that the flow is laminar — fluid layers slide smoothly past each other without mixing. The Reynolds number quantifies this: it is the ratio of inertial forces (which cause turbulence) to viscous forces (which suppress turbulence). Low Re means viscous forces dominate and flow is orderly.

Alternative Method — Dimensional Analysis for Stokes’ Law

We can verify Stokes’ law dimensionally. The drag force $F$ depends on $\eta$ , $r$ , and $v$ :

F = k \cdot \eta^a \cdot r^b \cdot v^c

Solving dimensions: $[MLT^{-2}] = [ML^{-1}T^{-1}]^a [L]^b [LT^{-1}]^c$

This gives $a = 1, b = 1, c = 1$ , so $F \propto \eta r v$ . The numerical factor $6\pi$ comes from solving the Navier-Stokes equation.

For NEET, terminal velocity $v_t \propto r^2$ . If the radius doubles, terminal velocity increases 4 times. Also, $v_t$ is inversely proportional to viscosity — a ball falls much slower through honey than water. These proportionalities are frequently tested.

Common Mistake

Students forget to account for buoyancy and use $\rho_s$ instead of $(\rho_s - \rho_f)$ in the terminal velocity formula. The effective weight is $mg - F_{buoyancy}$ , not just $mg$ . If the ball is lighter than the fluid ( $\rho_s < \rho_f$ ), the ball actually rises (negative terminal velocity), like a helium balloon in air.

Question

Solution — Step by Step

Why This Works

Alternative Method — Dimensional Analysis for Stokes’ Law

Common Mistake

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