Why does the viscosity of liquids decrease with temperature but that of gases increase?

In a liquid, viscosity arises from cohesive forces between molecules. Heating loosens these intermolecular bonds, so layers slide past one another more easily and viscosity falls. In a gas, viscosity arises from molecules carrying momentum between layers as they move randomly. Heating speeds up this random motion, increasing the momentum transport, so the gas becomes more viscous. NCERT states it plainly: the viscosity of liquids decreases with temperature, while it increases in the case of gases.

What is the difference between poise and poiseuille (Pa·s)?

Poiseuille (Pl), equal to N s m⁻² or Pa·s, is the SI unit of the coefficient of viscosity. Poise (P) is the CGS unit. They differ by a factor of ten: 1 poise = 0.1 Pa·s, so 1 Pa·s = 10 poise. Water at room temperature has a viscosity of about 1.0 × 10⁻³ Pa·s, which equals 1 centipoise.

Why is terminal velocity proportional to the square of the radius?

Weight minus buoyancy scales with volume, i.e. with r³, while the Stokes drag scales with r (since F = 6πηrv). Balancing the two at terminal velocity, the v that makes 6πηrv equal to (4/3)πr³(ρ−σ)g works out to v_t = (2/9)·r²(ρ−σ)g/η. The single power of r in the drag cancels one power in r³, leaving r². A drop with twice the radius therefore falls four times as fast at terminal velocity.

What are the assumptions behind Stokes' law?

Stokes' law F = 6πηrv assumes a small, smooth, rigid sphere moving slowly through an infinite, homogeneous, incompressible fluid, at low Reynolds number where the flow around the sphere stays streamline (laminar). It breaks down for large or fast bodies where the flow turns turbulent, near container walls, and for non-spherical bodies. NEET often plants a 'large sphere' or 'high speed' detail as a flag that Stokes' law no longer applies.

Does a heavier raindrop necessarily fall faster at terminal velocity?

Terminal velocity depends on radius and the density difference (ρ−σ), not on mass directly: v_t = (2/9)·r²(ρ−σ)g/η. A heavier drop is heavier because it is bigger, so its larger radius is what raises its terminal velocity. Two drops of the same radius but different mass cannot exist for a given liquid, since mass is fixed by radius and density. The controlling variable is r², so reason from radius, not from weight.

Viscosity & Stokes' Law — NEET Physics

Q: What is the velocity gradient and how does it relate to viscous force?

The velocity gradient dv/dx is the rate at which the layer speed changes with distance measured perpendicular to the flow. In the two-plate setup the speed rises uniformly from zero at the fixed plate to v at the moving plate, giving a constant gradient v/x. The viscous force obeys F = ηA(dv/dx): it is larger for a bigger contact area, a steeper velocity gradient, and a more viscous fluid.

What viscosity is

Most real fluids are not ideal; they resist motion. This resistance is an internal friction analogous to the friction felt when a solid slides over a surface, and it exists only when there is relative motion between adjacent layers of the fluid. NCERT §9.5 introduces it with a simple picture: a film of oil is enclosed between two glass plates, the bottom one fixed, the top one dragged sideways at a constant velocity $v$. Replace the oil with honey and a much larger force is needed for the same speed — honey is more viscous than oil.

The key experimental fact is that the fluid in contact with a surface moves with that surface. The layer touching the moving top plate travels at $v$; the layer touching the fixed bottom plate is at rest. Between them the layer speeds rise uniformly from zero to $v$. Each layer is dragged forward by the faster layer above it and held back by the slower layer below it, and this tug between layers is the viscous force. Because the layers slide smoothly over one another like the pages of a flat book pushed at the top cover, this orderly regime is called laminar flow.

Figure 1. A liquid film between two plates. The layer speed rises uniformly from zero at the fixed plate to $v$ at the moving plate; the slope of this profile is the velocity gradient $dv/dx$.

Velocity gradient and the coefficient η

Let $v$ be the speed of a layer at distance $x$ from the fixed plate and $v + dv$ the speed at $x + dx$. The quantity $dv/dx$ — the rate at which layer speed changes across the flow — is the velocity gradient. Experiment shows the tangential viscous force $F$ between two layers is proportional both to the contact area $A$ of the layers and to the velocity gradient:

$$ F = \eta\, A\, \frac{dv}{dx} $$

The constant of proportionality $\eta$ (pronounced "eta") is the coefficient of viscosity. NCERT phrases the same idea as a stress–strain-rate law: because the shear strain in a flowing fluid grows continuously with time, the relevant variable is the strain rate $v/l$, and $\eta$ is defined as the ratio of shearing stress to strain rate. A negative sign is sometimes written, $F = -\eta A\,(dv/dx)$, to mark that the force is frictional and opposes the relative motion.

Units of viscosity

From $F = \eta A\,(dv/dx)$, the dimensions of $\eta$ are $[\text{ML}^{-1}\text{T}^{-1}]$. The SI unit is named the poiseuille (Pl), equivalent to $\text{N s m}^{-2}$ or $\text{Pa·s}$. The CGS unit is the poise (P). The two are related by a factor of ten.

System	Unit	Equivalent	Note
SI	poiseuille (Pl)	`N s m⁻² = Pa·s`	Preferred NEET unit
CGS	poise (P)	`dyn s cm⁻²`	$1\text{ P} = 0.1\text{ Pa·s}$
Conversion	—	`1 Pa·s = 10 P`	So 1 Pa·s = 1000 cP
Dimensions	—	`[ML⁻¹T⁻¹]`	Same in every system

For scale, water at room temperature has $\eta \approx 1.0\times10^{-3}\ \text{Pa·s}$ (1 centipoise), blood is roughly $2.7\times10^{-3}\ \text{Pa·s}$ — "thicker" than water — and air is far thinner, around $1.8\times10^{-5}\ \text{Pa·s}$. NCERT also notes that the relative viscosity of blood (its viscosity divided by that of water) stays nearly constant between $0\,^{\circ}\text{C}$ and $37\,^{\circ}\text{C}$.

Temperature dependence

Viscosity changes with temperature, and the direction of change is opposite for liquids and gases. NCERT states it in one line: the viscosity of liquids decreases with temperature, while it increases in the case of gases. The reason lies in what carries the internal friction in each case.

Liquids — viscosity DECREASES on heating

Friction comes from cohesive intermolecular forces between layers.
Heating loosens these bonds, so layers slide past one another more easily.
Example: honey and glycerine flow much more freely when warmed.

Gases — viscosity INCREASES on heating

Friction comes from molecules carrying momentum between layers as they move randomly.
Heating speeds up that random motion, increasing the momentum transport between layers.
So a hotter gas resists shear more strongly.

Stokes' law and its assumptions

When a body falls through a fluid it drags the layer of fluid in contact with it, setting up relative motion between layers and so experiencing a retarding viscous force. Sir George G. Stokes (1819–1903) showed that for a small smooth sphere of radius $r$ moving slowly with velocity $v$ through a fluid of viscosity $\eta$, the drag is

$$ F = 6\pi\eta r v $$

This is Stokes' law. The drag is proportional to velocity and opposite to the direction of motion — an instructive example of a velocity-dependent retarding force. The dependence on $\eta$, $r$ and $v$ can be recovered by dimensional analysis: writing $F \propto \eta^a r^b v^c$ and matching dimensions gives $a=b=c=1$, so $F \propto \eta r v$; experiment fixes the numerical constant at $6\pi$.

Stokes' law is not universal. It holds only under a set of idealisations, and NEET routinely tests whether they apply.

Assumption	Why it matters
Sphere is small and rigid	The $6\pi r$ factor is derived for a smooth sphere; other shapes need different coefficients.
Speed is low (low Reynolds number)	Flow around the sphere must stay streamline/laminar; at high speed it turns turbulent and drag grows faster than $v$.
Fluid is infinite and homogeneous	Container walls or a finite column distort the flow and add wall drag.
Fluid is incompressible and continuous	The sphere must be large compared with the fluid's molecular spacing.

Forward link

Whether the flow stays laminar or turns turbulent is decided by a single dimensionless number — see Reynolds number for the onset of turbulence.

Terminal velocity derivation

Consider a sphere of radius $r$ and density $\rho$ released from rest in a fluid of density $\sigma$. Three forces act on it: its weight $W$ downward, the buoyant (upthrust) force $B$ upward, and the viscous drag $F$ upward (opposing the downward motion). At first the body accelerates, but as $v$ grows the drag grows with it. Eventually the net force becomes zero and the body falls at a constant terminal velocity $v_t$.

Figure 2. At terminal velocity the upward forces — buoyancy $B$ and Stokes drag $F$ — together balance the downward weight $W$. The net force, and hence the acceleration, is zero.

Write the three magnitudes for a sphere of volume $\tfrac{4}{3}\pi r^3$:

$$ W = \tfrac{4}{3}\pi r^3 \rho g, \qquad B = \tfrac{4}{3}\pi r^3 \sigma g, \qquad F = 6\pi\eta r v_t $$

At terminal velocity the net force is zero, so the upward drag and buoyancy together balance the weight:

$$ 6\pi\eta r v_t = \tfrac{4}{3}\pi r^3 \rho g - \tfrac{4}{3}\pi r^3 \sigma g = \tfrac{4}{3}\pi r^3 (\rho - \sigma) g $$

Cancelling $\pi r$ from both sides and solving for $v_t$:

$$ \boxed{\,v_t = \frac{2}{9}\,\frac{r^2 (\rho - \sigma) g}{\eta}\,} $$

Three features deserve emphasis. Terminal velocity goes as the square of the radius, so a drop with twice the radius falls four times as fast. It is inversely proportional to viscosity, so the body settles more slowly in a thicker medium. And it depends on the density difference $(\rho - \sigma)$: if the sphere is denser than the fluid it sinks at steady $v_t$; if it is less dense the bracket is negative and the body rises instead (an air bubble in water).

Applications — raindrops and parachutes

The terminal-velocity balance explains two everyday phenomena. Raindrops accelerate briefly as they begin to fall, but viscous drag in air quickly rises to meet the weight, and the drop reaches the ground at a modest constant speed rather than a lethal one — which is why falling rain does not injure. A parachutist exploits the same idea: a soldier leaving an aircraft first accelerates under gravity, but air drag reduces the acceleration until a terminal velocity is reached, and opening the parachute greatly increases the effective drag area so the descent is slow enough to land safely.

Worked examples

NCERT Example 9.9

The terminal velocity of a copper ball of radius 2.0 mm falling through a tank of oil at $20\,^{\circ}\text{C}$ is $6.5\ \text{cm s}^{-1}$. Compute the viscosity of the oil. Density of oil $=1.5\times10^{3}\ \text{kg m}^{-3}$, density of copper $=8.9\times10^{3}\ \text{kg m}^{-3}$, $g = 9.8\ \text{m s}^{-2}$.

Rearrange the terminal-velocity formula for $\eta$: $\eta = \dfrac{2}{9}\dfrac{r^2(\rho-\sigma)g}{v_t}$.

Substitute $r = 2\times10^{-3}\ \text{m}$, $v_t = 6.5\times10^{-2}\ \text{m s}^{-1}$, $(\rho-\sigma) = (8.9-1.5)\times10^{3} = 7.4\times10^{3}\ \text{kg m}^{-3}$: $\eta = \tfrac{2}{9}\cdot\dfrac{(2\times10^{-3})^2(7.4\times10^{3})(9.8)}{6.5\times10^{-2}} \approx 9.9\times10^{-1}\ \text{kg m}^{-1}\text{s}^{-1}$.

Answer: $\eta \approx 0.99\ \text{Pa·s}$.

NCERT Example 9.8

A metal block of area $0.10\ \text{m}^2$ rests on a liquid film of thickness $0.30\ \text{mm}$ on a table and is pulled by a string over a pulley carrying a $0.010\ \text{kg}$ mass. The block then moves at a constant $0.085\ \text{m s}^{-1}$. Find the coefficient of viscosity of the liquid.

Shear force. At constant speed the tension equals the hanging weight: $F = mg = 0.010\times9.8 = 9.8\times10^{-2}\ \text{N}$.

Apply $F = \eta A\,(v/x)$: the strain rate is $v/x = 0.085 / (0.30\times10^{-3}) = 283\ \text{s}^{-1}$, and shear stress is $F/A = 9.8\times10^{-2}/0.10 = 0.98\ \text{N m}^{-2}$.

Answer: $\eta = \dfrac{F/A}{v/x} = \dfrac{0.98}{283} \approx 3.46\times10^{-3}\ \text{Pa·s}$.

Quick recap

Viscosity and Stokes' law in one breath

Viscosity is internal friction between fluid layers; $F = \eta A\,(dv/dx)$, with the velocity gradient (not velocity) the key factor.
SI unit poiseuille (Pa·s); CGS unit poise; $1\ \text{Pa·s} = 10\ \text{P}$; dimensions $[\text{ML}^{-1}\text{T}^{-1}]$.
Heating: liquid viscosity decreases, gas viscosity increases — opposite directions, opposite mechanisms.
Stokes' drag on a small slow sphere: $F = 6\pi\eta r v$, valid only at low Reynolds number (laminar flow).
Force balance $6\pi\eta r v_t = \tfrac{4}{3}\pi r^3(\rho-\sigma)g$ gives $v_t = \tfrac{2}{9}\,r^2(\rho-\sigma)g/\eta$.
$v_t \propto r^2$ and $\propto 1/\eta$; power dissipated at terminal velocity $\propto r^5$. Raindrops and parachutes settle at $v_t$.

Viscosity & Stokes' Law