1. Physics
  2. Properties of Solids and Liquids
  3. Viscosity and Stokes' Law

Properties of Solids and Liquids · JEE & NEET Physics

Viscosity and Stokes' Law: notes and previous year questions

The friction inside a fluid: F = ηA dv/dy, streamline and turbulent flow, Stokes' law, terminal velocity and flow through pipes.

Viscosity and Stokes' Law in short

  • Viscosity is a fluid's internal friction between sliding layers.
  • F = ηA dv/dy; η is measured in Pa·s.
  • Liquids get less viscous when hot; gases get more viscous.
  • The Reynolds number decides between streamline and turbulent flow.

1Sticky fluids

Viscosity is a fluid's internal friction: it opposes layers sliding past each other. Water and air have low viscosity; oil, glycerin and honey high.

Liquids get runnier when warm (their molecules hold each other less). Gases get more viscous when hot (faster molecules carry more momentum between layers). Engine oil like 10W-30 is made to stay usable from cold starts to a hot engine; blood is 3–4 times as viscous as water.

2Layers sliding

Liquid touching a fixed plate is still; liquid touching a moving plate moves with it. In between, the speed changes steadily: the velocity gradient dv/dydv/dy.

F=ηAdvdyF = \eta A \frac{dv}{dy}η: coefficient of viscosity. Unit Pa·s; 1 poise = 0.1 Pa·s. Dimensions [M L⁻¹ T⁻¹].
Fluid (20 °C)η (Pa·s)
Air1.8×10−51.8\times10^{-5}
Water1.0×10−31.0\times10^{-3}
Blood33–4×10−34\times10^{-3}
Olive oil0.08
Glycerin1.5
Honey≈ 2–10

A 0.5 m² plate on 1 mm of oil (η=0.1\eta = 0.1 Pa·s) at 2 cm/s needs F=0.1×0.5×0.02÷0.001=1F = 0.1\times0.5\times0.02\div0.001 = 1 N.

3Streamline or turbulent

Slow flow moves in smooth layers (streamline or laminar flow); each particle follows a streamline. Streamlines never cross, and they crowd together where the flow is fastest. Fast flow breaks into eddies (turbulent flow).

Re=ρvDηRe = \frac{\rho v D}{\eta}Reynolds number, no unit. Roughly: below 1000 streamline, above 2000 turbulent.

Water in a 2 cm pipe: Re=20,000 vRe = 20{,}000\,v, so it reaches 2000 at 10 cm/s. Faster flow, a wider pipe or a denser liquid raise Re; more viscosity lowers it.

4Stokes' law

F=6πηrvF = 6\pi\eta r vDimensions force F = k η r v; Stokes found k = 6π. The drag opposes the motion.

It holds for a small, smooth sphere moving slowly (streamline flow round it) in a large body of fluid. For fast or large bodies the drag grows like v2v^2 instead. Example: 2 mm sphere, 5 cm/s, water: F≈1.9×10−6F \approx 1.9\times10^{-6} N.

5Terminal velocity

A ball dropped into a liquid speeds up until drag + buoyancy balance its weight; then the net force is zero and it falls at a steady terminal velocity:

43πr3ρg=43πr3σg+6πηrvt\tfrac43\pi r^3\rho g = \tfrac43\pi r^3\sigma g + 6\pi\eta r v_t
vt=2r2(ρ−σ)g9ηv_t = \frac{2r^2(\rho - \sigma)g}{9\eta}ρ: the sphere, σ: the liquid. v_t ∝ r².

Starting from rest, v=vt(1−e−t/τ)v = v_t(1 - e^{-t/\tau}) with τ=m/6πηr\tau = m/6\pi\eta r; it reaches 90% of vtv_t after about 2.3τ2.3\tau.

6Flow in pipes

Q=πr4(P1−P2)8ηLQ = \frac{\pi r^4 (P_1 - P_2)}{8\eta L}Poiseuille's law, for streamline flow. Fastest in the middle, still at the wall.

Q∝r4Q \propto r^4: half the radius gives 1/161/16 of the flow, and a radius only 16% smaller (84%) halves it, which is why narrowed arteries are dangerous. Q∝ΔPQ \propto \Delta P: 10 cm³/s at 20 Pa becomes 25 cm³/s at 50 Pa.

Summary

Key ideas

  • Viscosity is a fluid's internal friction between sliding layers.
  • F = ηA dv/dy; η is measured in Pa·s.
  • Liquids get less viscous when hot; gases get more viscous.
  • The Reynolds number decides between streamline and turbulent flow.
  • Stokes' law F = 6πηrv holds for small, slow spheres.
  • At terminal velocity the net force is zero; v_t ∝ r².
  • Flow through a pipe goes as r⁴ (Poiseuille).

Every equation

Viscous force
F=ηA dv/dyF = \eta A\,dv/dy
Poise
1 poise=0.1 Pa s1\ \text{poise} = 0.1\ \text{Pa s}
Reynolds number
Re=ρvD/ηRe = \rho v D/\eta
Stokes' law
F=6πηrvF = 6\pi\eta r v
Terminal velocity
vt=2r2(ρ−σ)g/9ηv_t = 2r^2(\rho-\sigma)g/9\eta
Approach to v_t
v=vt(1−e−t/τ), τ=m/6πηrv = v_t(1 - e^{-t/\tau}),\ \tau = m/6\pi\eta r
Poiseuille
Q=πr4ΔP/8ηLQ = \pi r^4\Delta P/8\eta L

Previous year questions with solutions

Real JEE and NEET questions on viscosity and stokes' law. Try each one before you open the solution.

Q1JEE Main 2026One correct option

If an air bubble of diameter 2 mm rises steadily through a liquid of density 2000 kg/m³ at a rate of 0.5 cm/s, then the coefficient of viscosity of liquid is ______ Poise. (Take g = 10 m/s²)

  1. A0.88
  2. B8.8
  3. C88.8
  4. D0.088
Show answer and solution

Answer: Option B

The bubble rises steadily, so it is at terminal velocity; air's density is ignored. With r=1 mmr = 1\ \mathrm{mm} (half the diameter) and v=0.5 cm/s=5×10−3 m/sv = 0.5\ \mathrm{cm/s} = 5 \times 10^{-3}\ \mathrm{m/s}, turn vt=2r2σg9ηv_{t} = \dfrac{2r^{2}\sigma g}{9\eta} round for η\eta:

η=2r2σg9v=2×10−6×2000×109×5×10−3=0.040.045≈0.89 Pa s\eta = \dfrac{2r^{2}\sigma g}{9v} = \dfrac{2 \times 10^{-6} \times 2000 \times 10}{9 \times 5 \times 10^{-3}} = \dfrac{0.04}{0.045} \approx 0.89\ \mathrm{Pa\,s}.

The question asks for poise, and 1 Pa s=101\ \mathrm{Pa\,s} = 10 poise, so η≈8.9\eta \approx 8.9 poise: option B, 8.88.8.

The trap is A, 0.880.88, the right number in the wrong unit: stopping at Pa s\mathrm{Pa\,s}. D divides by 1010 instead of multiplying. Using the diameter as rr would give four times too much, about 3636 poise.

Q2NEET 2023One correct option

The viscous drag acting on a metal sphere of diameter 1mm1 \mathrm{mm}, falling through a fluid of viscosity 0.8Pa0.8 \mathrm{Pa} s with a velocity of 2ms−12 ms^{-1} is equal to :

  1. A15×10−3N15\times {10}^{-3} N
  2. B30×10−3N30\times {10}^{-3} N
  3. C1.5×10−3N1.5\times {10}^{-3} N
  4. D20×10−3N20\times {10}^{-3} N
Show answer and solution

Answer: Option A

r=0.5 mm=5×10−4 mr = 0.5\ \mathrm{mm} = 5 \times 10^{-4}\ \mathrm{m}, and η=0.8 Pa s\eta = 0.8\ \mathrm{Pa\,s} is already in SI units. So

F=6πηrv=6×3.14×0.8×5×10−4×2≈1.5×10−2 N=15×10−3 NF = 6\pi\eta r v = 6 \times 3.14 \times 0.8 \times 5 \times 10^{-4} \times 2 \approx 1.5 \times 10^{-2}\ \mathrm{N} = 15 \times 10^{-3}\ \mathrm{N}.

The trap is C, 1.5×10−3 N1.5 \times 10^{-3}\ \mathrm{N}, which appears if the viscosity is shrunk by a factor of 1010, as though 0.8 Pa s0.8\ \mathrm{Pa\,s} needed converting (it does not; it is poise that would). B, 30×10−3 N30 \times 10^{-3}\ \mathrm{N}, uses the diameter as the radius.

Q3JEE Advanced 2018One or more correct options

Consider a thin square plate floating on a viscous liquid in a large tank. The height hh of the liquid in the tank is much less than the width of the tank. The floating plate is pulled horizontally with a constant velocity u0u_{0}. Which of the following statements is (are) true?

  1. AThe resistive force of liquid on the plate is inversely proportional to hh
  2. BThe resistive force of liquid on the plate is independent of the area of the plate
  3. CThe tangential (shear) stress on the floor of the tank increases with u0u_{0}
  4. DThe tangential (shear) stress on the plate varies linearly with the viscosity η\eta of the liquid
Show answer and solution

Answer: Options A, C, D

The liquid is shallow compared with the tank's width, so under the plate it is a thin film: at rest on the floor, moving at u0u_{0} with the plate, the speed rising steadily in between. The drag on the plate is F=ηAu0hF = \dfrac{\eta A u_{0}}{h} and the shear stress is ηu0h\dfrac{\eta u_{0}}{h}.

A: F∝1hF \propto \dfrac{1}{h}, true. B: F∝AF \propto A, so it does depend on the area; false. C: the stress is the same at every level of the film, so on the floor too it is ηu0h\dfrac{\eta u_{0}}{h}, which grows with u0u_{0}; true. D: the stress ηu0h\dfrac{\eta u_{0}}{h} is proportional to η\eta, a straight-line dependence; true. So A, C and D.

The trap is leaving out C, thinking the floor, being at rest, feels no stress. The floor holds back the bottom layer, and the steady film passes the same stress down to it that the plate applies at the top.

Practice questions, easy to hard

Three questions from the viscosity and stokes' law practice ladder: one easy, one medium, one hard.

Q4One correct option

From η=F/Adv/dx\eta = \dfrac{F/A}{dv/dx}, the SI unit of η\eta is N/m2s−1=N s m−2\dfrac{\mathrm{N/m^{2}}}{\mathrm{s^{-1}}} = \mathrm{N\,s\,m^{-2}}, which is the same as Pa s\mathrm{Pa\,s} (pascal second), sometimes called the decapoise. The older CGS unit, the poise, is ten times smaller:

1 poise=0.1 Pa s1\ \mathrm{poise} = 0.1\ \mathrm{Pa\,s}, so 1 Pa s=10 poise1\ \mathrm{Pa\,s} = 10\ \mathrm{poise}

Water at room temperature has η≈10−3 Pa s\eta \approx 10^{-3}\ \mathrm{Pa\,s}; honey's is thousands of times larger.

Glycerine's viscosity is about 1.5 Pa s1.5\ \mathrm{Pa\,s}. In poise, that is:

  1. A0.15 poise0.15\ \mathrm{poise}
  2. B15 poise15\ \mathrm{poise}
  3. C150 poise150\ \mathrm{poise}
  4. D1.5 poise1.5\ \mathrm{poise}
Show answer and solution

Answer: Option B

Each Pa s\mathrm{Pa\,s} is 1010 poise, so 1.5 Pa s=15 poise1.5\ \mathrm{Pa\,s} = 15\ \mathrm{poise}. The smaller unit needs the bigger number.

The trap is A, dividing by 1010 instead of multiplying: that treats the poise as the larger unit. C multiplies by 100100, and D forgets that the units differ at all. The prefix "deca" in decapoise is the reminder: one Pa s\mathrm{Pa\,s} is ten poise.

Q5Numerical answer

The density part of vtv_{t} is the difference (ρ−σ)(\rho - \sigma), not the sphere's density ρ\rho on its own: the upthrust takes its share before the drag has anything to balance.

An aluminium ball (density 2700 kg/m32700\ \mathrm{kg/m^{3}}) and a steel ball (density 7800 kg/m37800\ \mathrm{kg/m^{3}}) of the same radius fall through water (density 1000 kg/m31000\ \mathrm{kg/m^{3}}). The steel ball's terminal speed is how many times the aluminium ball's?

Show answer and solution

Answer: 4

Same radius, same liquid, so vsteelvAl=7800−10002700−1000=68001700=4\dfrac{v_{\mathrm{steel}}}{v_{\mathrm{Al}}} = \dfrac{7800 - 1000}{2700 - 1000} = \dfrac{6800}{1700} = 4.

The trap is 78002700≈2.9\dfrac{7800}{2700} \approx 2.9, the ratio of the densities themselves. The water cancels a larger fraction of the aluminium's weight than of the steel's, so the steel ball wins by more than its density alone suggests.

Q6Numerical answer

Energy gives a second route to the same power. Once a body falls at a steady speed its kinetic energy stops changing, and since the drag then equals weight −- upthrust, the power lost to drag is (W−U) vt(W - U)\,v_{t}, where WW is the weight and UU the upthrust. For a raindrop in air the upthrust is ignored (air's density is tiny next to water's), so the heat is produced at the rate mgvtmgv_{t}: all the energy gravity supplies is turned into heat.

A raindrop of mass 0.05 g0.05\ \mathrm{g} falls at a steady 8 m/s8\ \mathrm{m/s}. Taking g=10 m/s2g = 10\ \mathrm{m/s^{2}}, at what rate is heat produced, in milliwatts?

Show answer and solution

Answer: 4 mW

m=5×10−5 kgm = 5 \times 10^{-5}\ \mathrm{kg}, so mgvt=5×10−5×10×8=4×10−3 W=4 mWmgv_{t} = 5 \times 10^{-5} \times 10 \times 8 = 4 \times 10^{-3}\ \mathrm{W} = 4\ \mathrm{mW}.

The trap is reaching for the kinetic energy, 12mv2\dfrac{1}{2}mv^{2}: at a steady speed that does not change, so none of the heat comes from it. Every second the drop falls 8 m8\ \mathrm{m}, and the potential energy it loses over those 8 m8\ \mathrm{m} becomes heat.