Viscosity and Stokes' Law

What Viscosity Actually Means

Imagine pouring honey from a bottle and pouring water from the same bottle — the honey takes forever, the water rushes out. That stickiness, that resistance to flow, is what we call viscosity. In physics, it’s a property of fluids that measures how much internal friction there is between adjacent layers when they slide past each other.

For JEE and NEET, viscosity shows up in two forms: as a coefficient in Newton’s law of viscous flow, and as the drag force on a sphere falling through a fluid (Stokes’ law). Both are scoring topics in the fluid mechanics weightage — typically $1$ - $2$ marks on JEE Main and a guaranteed concept question on NEET.

We’ll work through the definitions, derive the formulas you need, solve a graded set of problems, and close with the traps that catch students. By the end of this hub, terminal velocity, Poiseuille flow, and viscosity-from-graph problems should all feel routine.

Key Terms & Definitions

Viscosity ( $\eta$ ): The constant of proportionality between the viscous shear stress and the velocity gradient in a fluid. SI unit: $\text{Pa·s}$ (or $\text{N·s/m}^2$ ). CGS unit: poise. $1 \text{ Pa·s} = 10 \text{ poise}$ .

Velocity gradient ( $dv/dy$ ): How quickly fluid speed changes as you move perpendicular to the flow. Higher gradient = more shearing.

Reynolds number ( $Re$ ): A dimensionless number that tells you whether flow is laminar (smooth, low $Re$ ) or turbulent (chaotic, high $Re$ ). For a sphere falling through fluid, Stokes’ law assumes laminar flow.

Terminal velocity ( $v_t$ ): The constant velocity reached by a falling object when net force becomes zero — gravity is balanced by buoyancy plus viscous drag.

Streamline flow: Each fluid particle follows the same path as previous particles. Stokes’ law and Poiseuille’s law assume streamline flow.

Methods/Concepts

Newton’s Law of Viscosity

When a fluid flows in layers, the viscous force between two adjacent layers of area $A$ , separated by a small distance $dy$ , with relative velocity $dv$ , is:

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

The minus sign (when written formally) reflects that viscous force opposes relative motion. Units check: $[\eta] = [F]/([A][dv/dy]) = \text{N·s/m}^2$ .

Stokes’ Law

A small sphere of radius $r$ moving through a fluid of viscosity $\eta$ at velocity $v$ experiences a drag force:

$F_d = 6\pi\eta r v$

This holds for low Reynolds numbers (slow motion or small spheres or high viscosity).

Terminal Velocity Derivation

A sphere of density $\rho_s$ and radius $r$ falling through a fluid of density $\rho_f$ experiences three forces:

Weight downward: $W = \tfrac{4}{3}\pi r^3 \rho_s g$
Buoyancy upward: $B = \tfrac{4}{3}\pi r^3 \rho_f g$
Viscous drag upward: $F_d = 6\pi\eta r v$

At terminal velocity, the net force is zero: $W = B + F_d$ . Solving:

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

Notice the $r^2$ dependence — a sphere with twice the radius reaches four times the terminal velocity. This is why dust takes ages to settle but pebbles drop fast.

Solved Examples

Example 1 (CBSE, easy)

A steel ball of radius $1 \text{ mm}$ and density $8000 \text{ kg/m}^3$ falls through glycerine of density $1300 \text{ kg/m}^3$ and viscosity $0.83 \text{ Pa·s}$ . Find the terminal velocity. ( $g = 10 \text{ m/s}^2$ .)

$v_t = \frac{2 \times (10^{-3})^2 \times (8000 - 1300) \times 10}{9 \times 0.83} = \frac{2 \times 10^{-6} \times 6700 \times 10}{7.47} \approx 1.79 \times 10^{-2} \text{ m/s}$

About $1.8 \text{ cm/s}$ — slow, as expected for glycerine.

Example 2 (JEE Main, medium)

The terminal velocity of a small spherical droplet of radius $r$ falling through air is $v_t$ . If two such droplets coalesce into one, what is the new terminal velocity?

When two droplets combine, volume doubles. So the new radius $r'$ satisfies $\tfrac{4}{3}\pi r'^3 = 2 \times \tfrac{4}{3}\pi r^3 \implies r' = 2^{1/3} r$ .

Since $v_t \propto r^2$ , the new terminal velocity is:

$v_t' = (2^{1/3})^2 v_t = 2^{2/3} v_t \approx 1.587 v_t$

This appeared in JEE Main 2023 in slightly different framing — droplets merging is a favourite JEE setup.

Example 3 (JEE Advanced, hard)

Two solid spheres of radii $R_1$ and $R_2$ but the same density fall through the same fluid. They reach terminal velocities $v_1$ and $v_2$ respectively. Show that the ratio $v_1/v_2 = (R_1/R_2)^2$ and use this to find $R_1/R_2$ given $v_1 = 4v_2$ .

Since both spheres have the same density and fluid, the terminal velocity formula reduces to $v_t \propto r^2$ . So $v_1/v_2 = (R_1/R_2)^2$ .

If $v_1 = 4 v_2$ , then $(R_1/R_2)^2 = 4 \implies R_1/R_2 = 2$ .

Exam-Specific Tips

JEE Main weightage: Fluid mechanics overall is $\sim 3\%$ — usually one question per shift. Viscosity and terminal velocity together account for a guaranteed concept question. Stokes’ law is also a feeder topic for advanced electrostatics (Millikan’s oil drop experiment).

NEET weightage: Fluid mechanics is part of “Properties of Bulk Matter” — usually $1$ - $2$ questions. NEET often tests the formula directly with droplet-coalescence twists.

CBSE Class 11: Numerical problems on terminal velocity and viscosity coefficient are standard. Derivations of Stokes’ law are not in the CBSE syllabus, but using the formula is.

For JEE multiple-choice, look for proportionality questions (“if radius doubles, terminal velocity becomes…”) rather than full numerical calculations. The $r^2$ dependence is the most-asked relationship.

Common Mistakes to Avoid

1. Forgetting buoyancy. Many students write $W = F_d$ at terminal velocity, ignoring the upward buoyant force. Always include all three: weight, buoyancy, drag.

2. Wrong sign for $\rho_s - \rho_f$ . If the sphere is less dense than the fluid, it rises, not falls. The formula still applies; the velocity is just upward. Check the sign of $\rho_s - \rho_f$ before plugging in.

3. Using diameter instead of radius. Stokes’ law uses radius, not diameter. A factor of $2$ here doubles your terminal velocity through the $r^2$ — a common multiple-choice trap.

4. Mixing units. Glycerine viscosity is often quoted in poise (CGS) — $1 \text{ poise} = 0.1 \text{ Pa·s}$ . Convert before plugging in.

5. Applying Stokes’ law in turbulent flow. Stokes’ law is valid only at low Reynolds number. For raindrops larger than a few mm, the drag is no longer linear in $v$ — JEE Advanced occasionally hints at this with an inequality on radius.

Practice Questions

Q1. A spherical raindrop of radius $0.5 \text{ mm}$ falls through air ( $\eta_{\text{air}} = 1.8 \times 10^{-5} \text{ Pa·s}$ , $\rho_{\text{air}} \approx 1.2 \text{ kg/m}^3$ , $\rho_{\text{water}} = 1000 \text{ kg/m}^3$ ). Find the terminal velocity.

$v_t = \dfrac{2(0.5 \times 10^{-3})^2(1000-1.2)(10)}{9 \times 1.8 \times 10^{-5}} \approx 30.8 \text{ m/s}$ . Note: real raindrops fall at $\sim 9 \text{ m/s}$ — Stokes’ law overestimates here because flow is turbulent at this radius.

Q2. Eight identical drops of mercury, each falling at terminal velocity $v$ , coalesce to form a single drop. What is the new terminal velocity?

New radius $r' = 8^{1/3} r = 2r$ . Since $v_t \propto r^2$ , new $v_t' = 4v$ .

Q3. The viscosity of a liquid is $0.6 \text{ Pa·s}$ . Find the viscous force per unit area when the velocity gradient is $50 \text{ s}^{-1}$ .

$F/A = \eta(dv/dy) = 0.6 \times 50 = 30 \text{ N/m}^2$ .

Q4. A small sphere reaches terminal velocity in oil. If the oil is replaced by another oil of double the viscosity (everything else same), how does the terminal velocity change?

$v_t \propto 1/\eta$ , so terminal velocity halves.

Q5. A sphere falls through a fluid. Plot velocity versus time qualitatively.

Velocity increases from zero, asymptotically approaches terminal velocity as drag grows to balance net gravity. The curve is concave down.

Q6. Why do small droplets fall slower than large ones from the same height?

Smaller radius means smaller terminal velocity ( $\propto r^2$ ). Both reach terminal velocity quickly; the smaller one settles at a slower rate.

Q7. What are the dimensions of $\eta$ ?

$[\eta] = ML^{-1}T^{-1}$ .

Q8. A bubble of air rises through water. Apply the terminal velocity formula and explain the sign.

$\rho_s - \rho_f$ is negative (air less dense than water), so $v_t$ is negative — meaning velocity is upward. Magnitude is the upward terminal speed.

FAQs

Q. Is viscosity the same as friction?

In a sense — viscosity is internal friction within a fluid. But unlike solid friction, it depends on velocity gradient, not just normal force.

Q. Does viscosity depend on temperature?

Yes. For liquids, viscosity decreases with temperature (warm honey flows easier). For gases, it increases with temperature (hotter molecules collide more). This sign difference often appears in NEET concept questions.

Q. When does Stokes’ law fail?

When Reynolds number gets large ( $Re > \sim 1$ for a sphere). For raindrops larger than $\sim 1 \text{ mm}$ in air, drag becomes quadratic in velocity instead of linear, and you need empirical drag coefficients.

Q. Why is the terminal velocity formula derived assuming a small sphere?

Because Stokes’ law itself requires laminar flow around the sphere, which only holds at small radius (or low velocity).

Q. What’s the difference between viscosity and surface tension?

Viscosity is about resistance to flow between layers. Surface tension is about energy at the boundary of a fluid. Different physics, both governed by molecular interactions.

Q. Can viscosity be negative?

No — viscosity is always positive. Negative viscosity would mean fluid layers spontaneously slide faster past each other, violating thermodynamics.