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JEE Mains Physics · Mechanical Properties of Fluids

Viscosity, Stokes' Law and Reynolds Number

Viscosity is a fluid's internal friction: the force between layers is ηA times the rate at which the speed changes across them, a small sphere moving slowly feels a drag of 6πηrv, and the Reynolds number ρvd/η says whether a flow stays smooth.

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Why this matters

Twelve PYQs, eight of them multiple choice, and none from 2026. Five use Newton's law of viscosity, the Reynolds number or the facts of viscous flow, such as how viscosity changes with temperature; seven ask for the viscous force on a ball falling at constant velocity, and three of those are the same question set in three different years.

Concept 1 of 2: Newton's law of viscosity and the Reynolds number

In a flowing liquid each layer drags on the next. The drag grows with the area of contact and with how quickly the speed changes from one layer to the next. In a thin film between a still surface and a moving plate, the speed goes from 0 to v across the thickness d, so the rate of change is v/d. The Reynolds number compares the push of the moving fluid with this drag: when it is large, the flow breaks up into turbulence.

Definition

  • F=ηAdvdzF = \eta A \dfrac{dv}{dz}. For a film or a layer of depth d: F=ηAvdF = \eta A\dfrac{v}{d}, and the shear stress is FA=ηvd\dfrac{F}{A} = \eta\dfrac{v}{d}.
  • Unit of η: Pa s. 1 Pa s = 10 poise.
  • Liquids: η FALLS as the temperature rises. Gases: η RISES with temperature. Gases are far less viscous than liquids.
  • Re=ρvdηR_e = \dfrac{\rho v d}{\eta}, with d the DIAMETER of the pipe. NCERT's bands: below about 1000 the flow is streamline, between 1000 and 2000 it is unsteady, above 2000 it is turbulent.
  • Flow from a tap: v=Qπr2v = \dfrac{Q}{\pi r^{2}}, with Q in m³/s (1 L/min = 160×10−3 m3/s\tfrac{1}{60} \times 10^{-3}\ \text{m}^{3}/\text{s}).
  • In steady flow two streamlines never cross.

Viscous force and Reynolds number

F=ηAvd,Re=ρvdηF = \eta A\frac{v}{d}, \qquad R_e = \frac{\rho v d}{\eta}

Worked example

A plate of area 0.5 m20.5\ \text{m}^{2} slides at 0.4 m/s over a film of oil 1 mm thick on a fixed table. The oil's viscosity is 0.2 Pa s. What force keeps the plate moving at this speed?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2022 · JEE Mains 2022 — 25 June · Q23Moderate

Example 1 · Mechanical Properties of Fluids · Viscosity, Stokes' Law and Reynolds Number

The velocity of upper layer of water in a river is 36kmh−136kmh^{- 1}. Shearing stress between horizontal layers of water is 10−3Nm−210^{- 3}Nm^{- 2}. Depth of the river is ____\_\_\_\_ m. (Coefficient of viscosity of water is  10−2Pa.s)\left. \ 10^{- 2}Pa.s \right)

Liquids and gases go opposite ways with temperature

Heating a liquid lowers its viscosity, so hot water flows faster. Heating a gas raises its viscosity. A statement that says viscosity rises with temperature is true only for a gas.

Convert km/h and poise first

72 km/h is 20 m/s, and 1 poise is 0.1 Pa s. A stem that mixes units gives an answer off by a power of ten.
Drill 4 more on newton's law of viscosity and the reynolds number

Concept 2 of 2: Stokes' law and the viscous force at constant velocity

A small sphere moving slowly through a fluid feels a drag of 6πηrv6\pi\eta r v. When a ball falls at constant velocity, nothing accelerates it, so the forces on it balance. The drag is then whatever is left of the weight after buoyancy. That gives the viscous force without knowing η or v.

Definition

  • Stokes' law: F=6πηrvF = 6\pi\eta r v, for a small sphere in slow streamline flow.
  • At constant velocity: Fv=mg−FBF_v = mg - F_B. The ball's volume is m/ρm/\rho, so FB=mρρ0gF_B = \dfrac{m}{\rho}\rho_0 g and Fv=mg(1−ρ0ρ)F_v = mg\left(1 - \dfrac{\rho_0}{\rho}\right).
  • If the ball is twice as dense as the liquid, Fv=mg/2F_v = mg/2. If buoyancy is neglected, Fv=mgF_v = mg.
  • From the radius and densities: Fv=43πr3(ρ−ρ0)gF_v = \tfrac{4}{3}\pi r^{3}(\rho - \rho_0)g.

Viscous force at constant velocity

Fv=mg(1−ρ0ρ)=43πr3(ρ−ρ0)gF_v = mg\left(1 - \frac{\rho_0}{\rho}\right) = \tfrac{4}{3}\pi r^{3}(\rho - \rho_0)g

Worked example

A steel ball of mass 4 g and density 7500 kg/m37500\ \text{kg/m}^{3} falls at constant velocity through oil of density 1500 kg/m31500\ \text{kg/m}^{3}. Find the viscous force on it. (g=10 m/s2g = 10\ \text{m/s}^{2})
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

JEE Mains · 2024 · 6 April 2024 · Q19Moderate

Example 2 · Mechanical Properties of Fluids · Viscosity, Stokes' Law and Reynolds Number

A small ball of mass mm and density ρ\rho is dropped in a viscous liquid of density ρ0\rho_{0}. After sometime, the ball falls with constant velocity. The viscous force on the ball is :

Buoyancy is subtracted, not added

Weight = buoyancy + viscous drag, so the drag is the weight MINUS the buoyancy. Adding them gives a force bigger than the weight.

The ball's density goes in the denominator

F = mg(1 − ρ_liquid/ρ_ball). The option mg(ρ_liquid/ρ_ball − 1) is negative for any ball that sinks.
Drill 6 more on stokes' law and the viscous force at constant velocity

Summary — formulas & gotchas at a glance

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Formulas (2)

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