Properties of Fluids

Q: Which properties of fluids are important for NEET?

NEET regularly asks surface energy and work in blowing bubbles, excess pressure inside drops and soap bubbles, capillary rise with Jurin's law, the angle of contact, and terminal velocity from Stokes' law. Most are ratio questions, such as how the rise changes when the tube radius doubles.

Q: What is surface tension and what is its SI unit?

Surface tension is the force per unit length acting along a liquid surface, or the energy needed per unit increase in surface area. Its SI unit is newton per metre, which is the same as joule per square metre.

Q: Why are raindrops and soap bubbles spherical?

Surface tension makes a liquid surface behave like a stretched membrane that tries to reach the least possible area. For a given volume a sphere has the smallest surface area, so free drops and bubbles take a spherical shape.

Q: Why is the excess pressure in a soap bubble 4T/R but in a drop only 2T/R?

A liquid drop has one surface, but a soap bubble is a thin film with an inner and an outer surface. Each surface contributes 2T/R, so the bubble's excess pressure is twice that of a drop of the same radius.

Q: Why does water rise in a capillary tube but mercury falls?

Water wets glass because adhesion exceeds cohesion, giving an angle of contact below 90 degrees and a concave meniscus, so it rises. Mercury does not wet glass, its angle of contact is about 140 degrees, and its level is pushed down.

Q: What is terminal velocity?

It is the constant speed reached by a body falling through a fluid when its weight is exactly balanced by the buoyant force and the viscous drag. For a small sphere it equals 2 r squared times the density difference times g, divided by 9 eta.

Q: What surface tension and viscosity questions come in JEE Main and Advanced?

JEE Main asks energy released when drops merge, excess pressure in bubbles and capillary rise. JEE Advanced adds joined bubbles, capillaries shorter than the rise height, capillaries in accelerating lifts, charged bubbles, terminal velocity graphs and viscous force between plates.

Q: What does the Reynolds number tell us?

The Reynolds number, rho v d over eta, compares inertial forces with viscous forces in a flow. In a pipe the flow is laminar when it is below about 1000 and turbulent when it is above about 2000.

PhysicsProperties Of Solids And FluidsFor NEET aspirants

The properties of fluids that go beyond pressure are surface tension and viscosity. Surface tension ( $T$ ) makes drops round, lifts water up thin tubes and sets the excess pressure inside bubbles; viscosity ( $η$ ) is fluid friction, giving Stokes' drag and terminal velocity. These properties of fluids are a steady source of NEET and JEE Main questions on capillary rise, bubbles and terminal velocity, all worked out below.

On this page1Recap2Surface tension3Drops and bubbles4Excess pressure5Angle of contact6Capillary rise7Viscosity8Stokes and terminal velocity9Reynolds number

Key Formulas - Quick Reference

★ Must learnSurface tension $T = \frac{F}{L}$ ( $N m^{- 1}$ ) = surface energy per unit area ( $J m^{- 2}$ )
Angle of contact: $cos θ = \frac{S _{s a} - S _{s l}}{S _{l a}}$ (NCERT)
Work to increase area: $W = T Δ A$ (film or bubble: $Δ A$ counts both surfaces)
★ Must learnExcess pressure: drop / air bubble in liquid $\frac{2 T}{R}$ ; soap bubble $\frac{4 T}{R}$
★ Must learnCapillary rise (Jurin's law): $h = \frac{2 T cos θ}{r ρ g}$
$n$ drops of radius $r$ merge into one of radius $R = n^{1/3} r$ ; energy released $= 4 π T (n r^{2} - R^{2})$
Newton's law of viscosity: $F = η A \frac{d v}{d x}$ ; $1 Pa s = 10 poise$
★ Must learnStokes' law: $F = 6 π η r v$
Poiseuille: flow rate $\frac{d Q}{d t} = \frac{π P r ^{4}}{8 η L}$
★ Must learnTerminal velocity: $v_{t} = \frac{2 r ^{2} ( ρ - σ ) g}{9 η}$
Reynolds number: $R_{e} = \frac{ρ v d}{η}$ ; laminar below about $1000$ , turbulent above about $2000$

1. Quick Recap: Pressure and Flow

A fluid is a substance that flows. Its pressure and flow laws are developed in full in Hydrostatics and Hydrodynamic; this page uses only their results.

Idea	Result	Where it is derived
Pressure at depth	$P = P_{0} + ρ g h$	Hydrostatics
Pascal's law	$\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}}$ (hydraulic lift, brakes)	Hydrostatics
Buoyancy	$F_{B} = ρ_{L} V_{in} g$	Hydrostatics
Continuity	$A_{1} v_{1} = A_{2} v_{2}$	Hydrodynamic
Bernoulli	$P + ρ g h + \frac{1}{2} ρ v^{2} = constant$	Hydrodynamic

Bernoulli's equation assumes a non-viscous fluid. Real fluids have viscosity (Section 3), and every free liquid surface has surface tension (Section 2).

2. Surface Tension

2.1 Molecular origin

The attraction between molecules of the same substance is cohesion; attraction between molecules of different substances (say water and glass) is adhesion. Cohesion is very strong in solids, so they have a definite shape and size; it is weaker in liquids, which keep a definite volume but take the shape of the container; it is negligible in gases, which have neither a fixed shape nor a fixed volume.

Every molecule attracts its neighbours within a short range (the dashed circles below). A molecule deep inside is pulled equally from all sides. A molecule at or near the surface has liquid only below it, so it feels a net inward pull. To bring a molecule to the surface, work must be done against this pull, so surface molecules have extra energy. A liquid therefore tends to keep its surface area as small as possible, which is why free drops and bubbles are spherical (a sphere has the least area for a given volume).

Figure 1: Surface molecules have no liquid above them, so they feel a net inward pull and carry extra energy. The surface therefore tends to shrink to the least area.

2.2 Definition: force per length and energy per area

Surface tension $T$ is the force per unit length acting in the plane of the surface, perpendicular to any line drawn on it: $T = \frac{F}{L}$ . Equivalently, it is the surface energy, the extra energy (work done) per unit increase in surface area: $T = \frac{W}{Δ A}$ .

SI unit $N m^{- 1}$ (same as $J m^{- 2}$ ); dimensions $[MT^{- 2}]$ . Water at $2 0^{\circ} C$ : $T \approx 0.073 N m^{- 1}$ .

To see why both definitions give the same number, take a soap film on a U-shaped wire frame with a sliding wire of length $L$ . The film has two surfaces, front and back, so it pulls the slider with $F = 2 T L$ . Moving the slider out by $Δ x$ does work $F Δ x = 2 T L Δ x = T Δ A$ , where $Δ A = 2 L Δ x$ is the new area of both surfaces.

Figure 2: A film has two surfaces, so the slider needs

F = 2 T L

. Pulling it out by

Δ x

does work

T \cdot 2 L Δ x

: surface tension = surface energy per unit area.

2.3 Work in forming drops and bubbles

Forming a surface costs energy $T \times$ (new area).

Process	New area	Work done / energy
Liquid drop of radius $R$	$4 π R^{2}$	$4 π R^{2} T$
Soap bubble of radius $R$ (two surfaces)	$2 \times 4 π R^{2}$	$8 π R^{2} T$
Big drop (radius $R$ ) broken into $n$ drops of radius $r$	$4 π (n r^{2} - R^{2})$	Energy must be supplied
$n$ small drops merge into one	area decreases	Energy $4 π T (n r^{2} - R^{2})$ is released (as heat)

Volume is conserved when drops split or merge: $n \cdot \frac{4}{3} π r^{3} = \frac{4}{3} π R^{3}$ , so $R = n^{1/3} r$ .

2.4 Factors that change surface tension

Temperature: surface tension falls as temperature rises and becomes zero at the critical temperature. Ferguson's relation: $T = T_{0} (1 - \frac{θ}{θ _{c}})^{n}$ , where $T_{0}$ is the surface tension at $0^{\circ} C$ , $θ$ the temperature, $θ_{c}$ the critical temperature and $n \approx 1.21$ (varies slightly between liquids). Hot soapy water cleans better partly for this reason.
Impurities: highly soluble substances such as $NaCl$ and $ZnS O_{4}$ raise the surface tension of water slightly; sparingly soluble substances such as phenol, soap and detergents lower it sharply.
Detergents in washing: detergent molecules sit with one end in water and the other on the grease, lowering the water-grease surface tension so that water can lift grease away in small drops.

2.5 Excess pressure across a curved surface

Consider a molecule on a liquid surface and the pull of its neighbours in the surface.

Surface	Resultant pull of surface tension	Pressure difference
Plane	Zero: pulled equally in all directions	None: equal pressure on both sides
Concave (curving up)	Upward, towards the centre of curvature	Pressure below (concave side) exceeds pressure above by $\frac{2 T}{R}$
Convex (curving down)	Downward, towards the centre of curvature	Pressure below (concave side, inside) exceeds pressure above by $\frac{2 T}{R}$

So there is always an excess of pressure on the concave side of a curved liquid surface. Energy method for a drop of radius $R$ with inside pressure $P_{i}$ and outside $P_{0}$ :

Let the radius grow by $d R$ against the excess pressure. Work done by the pressure excess: $(P_{i} - P_{0}) \cdot 4 π R^{2} d R$ .
Increase in surface energy: $T \cdot d (4 π R^{2}) = T \cdot 8 π R d R$ .
Equate:
$P_{i} - P_{0} = \frac{2 T}{R} (drop, or air bubble inside a liquid)$
A soap bubble in air has two surfaces, so the surface energy term doubles:
$P_{i} - P_{0} = \frac{4 T}{R} (soap bubble)$

Force method (same result). Cut a soap bubble of radius $r$ into two halves and look at one half:

Surface tension acts along the cut rim on both the inner and the outer surface: $F_{T} = T \times 2 (2 π r) = 4 π r T$ .
The excess pressure pushes the half outward over its projected area $π r^{2}$ : $F_{P} = (p - p_{a}) π r^{2}$ .
Equilibrium: $(p - p_{a}) π r^{2} = 4 π r T$ , so $p - p_{a} = \frac{4 T}{r}$ . For a drop (one surface) the rim force is $2 π r T$ , giving $\frac{2 T}{r}$ .

Figure 3: Cut a soap bubble in half. The excess pressure pushes the half outward over the projected area

π R^{2}

; surface tension pulls back along the cut rim on both surfaces,

2 \times T \times 2 π R

. So

(P_{i} - P_{o}) π R^{2} = 4 π R T

, giving

\frac{4 T}{R}

(and

\frac{2 T}{R}

for one surface).

Figure 4: Pressure is always higher on the concave side (

P_{i}

inside,

P_{o}

just outside). One surface:

Δ P = \frac{2 T}{R}

; soap bubble (two surfaces):

Δ P = \frac{4 T}{R}

Exam Trick

Smaller bubble, larger pressure. $Δ P \propto \frac{1}{R}$ . If two soap bubbles of different sizes are joined by a tube, air flows from the smaller bubble into the bigger one: the small one shrinks and the big one grows. Count surfaces: one surface gives $2 T / R$ , two surfaces give $4 T / R$ .

Figure 5: Excess pressure

\propto \frac{1}{R}

. With

T = 0.03 N m^{- 1}

, a

1 cm

bubble has

\frac{4 T}{r} = 12 Pa

and a

2 cm

bubble only

6 Pa

, so air flows from the smaller bubble into the larger one until the small one is gone (dashed: sizes later).

JEE Advanced

Charged soap bubble. Charge on a bubble spreads over its surface with density $σ$ and pushes outward with an electrostatic pressure $\frac{σ ^{2}}{2 ε _{0}}$ , which helps the gas inside. The gas pressure needed for equilibrium therefore falls: before charging $p_{1} = p_{a} + \frac{4 T}{r _{1}}$ ; after charging $p_{2} = p_{a} + \frac{4 T}{r _{2}} - \frac{σ ^{2}}{2 ε _{0}}$ . At constant temperature Boyle's law ( $p_{1} V_{1} = p_{2} V_{2}$ ) gives

(p_{a} + \frac{4 T}{r _{1}}) r_{1}^{3} = (p_{a} + \frac{4 T}{r _{2}} - \frac{σ ^{2}}{2 ε _{0}}) r_{2}^{3}

from which $r_{2}$ can be found. The gas pressure falls, so its volume grows: a charged bubble expands, $r_{2} > r_{1}$ , whatever the sign of the charge.

2.6 Angle of contact

Where a liquid meets a solid, the liquid surface makes an angle $θ$ with the solid, measured inside the liquid. It depends on the balance between adhesion (liquid-solid attraction) and cohesion (liquid-liquid attraction).

Angle of contact	Behaviour	Examples
$θ < 9 0^{\circ}$	Liquid wets the solid, spreads; meniscus in a tube is concave	Water on ordinary glass ( $θ \approx 8^{\circ}$ ; taken as $0^{\circ}$ for pure water on clean glass), kerosene on glass
$θ > 9 0^{\circ}$	Liquid does not wet, forms beads; meniscus is convex	Mercury on glass ( $θ \approx 14 0^{\circ}$ ), water on a waxy or waterproofed surface

Figure 6: The angle of contact

θ

is measured inside the liquid between the solid surface and the tangent to the liquid surface. Drawn for

θ = 4 0^{\circ}

(water on glass, illustrative) and

θ = 14 0^{\circ}

(mercury).

At the contact line three surface tensions meet: $S_{l a}$ (liquid-air) along the tangent to the liquid surface, $S_{s l}$ (solid-liquid) along the solid towards the liquid, and $S_{s a}$ (solid-air) along the solid away from it. Balancing them along the solid surface (NCERT):

S_{l a} cos θ + S_{s l} = S_{s a} \Rightarrow cos θ = \frac{S _{s a} - S _{s l}}{S _{l a}}

If $S_{s l} < S_{s a}$ , $cos θ > 0$ and $θ$ is acute (the liquid wets); if $S_{s l} > S_{s a}$ , $θ$ is obtuse.

Figure 7: At the contact line the horizontal pulls balance (NCERT):

S_{l a} cos θ + S_{s l} = S_{s a}

, so

cos θ = \frac{S _{s a} - S _{s l}}{S _{l a}}

. If

S_{s l} < S_{s a}

θ

is acute and the liquid wets the solid; if

S_{s l} > S_{s a}

θ

is obtuse. Arrow lengths drawn to satisfy the balance.

Waterproofing agents increase $θ$ so water forms beads and rolls off; soaps and detergents decrease $θ$ so water spreads and soaks into fabric.

2.7 Capillary rise (Jurin's law)

A liquid that wets glass ( $θ < 9 0^{\circ}$ ) rises in a narrow tube dipped in it. The concave meniscus of radius $R$ has lower pressure just below it than the air above.

For a tube of radius $r$ , the meniscus radius is $R = \frac{r}{cos θ}$ .
Pressure just below the meniscus: $P_{0} - \frac{2 T}{R} = P_{0} - \frac{2 T cos θ}{r}$ .
This point is a height $h$ above the outside liquid surface (pressure $P_{0}$ ), so $P_{0} - \frac{2 T cos θ}{r} + ρ g h = P_{0}$ , giving
$h = \frac{2 T cos θ}{r ρ g}$

Figure 8: Jurin's law

h = \frac{2 T cos θ}{r ρ g}

. Heights drawn to scale:

h \propto \frac{1}{r}

. For mercury

θ > 9 0^{\circ}

, so

cos θ < 0

and the level falls.

Force method. Around the rim of the meniscus (length $2 π r$ ) surface tension pulls the liquid along the meniscus, at $θ$ to the wall. Its vertical part holds up the raised column: $2 π r T cos θ = π r^{2} h ρ g$ , which gives the same $h = \frac{2 T cos θ}{r ρ g}$ .

Figure 9: (a) Upward pull of surface tension around the rim

= 2 π r T cos θ

= weight of the column

π r^{2} h ρ g

, which gives Jurin's law. (b) If the tube sticks out only

h^{'} < h

, the liquid reaches the top and the meniscus flattens (

R^{'} = \frac{h R}{h ^{'}}

; drawn for

h^{'} = 0.6 h

θ

from

2 0^{\circ}

5 6^{\circ}

). The liquid never overflows.

For $θ > 9 0^{\circ}$ (mercury in glass) $cos θ < 0$ , so $h$ is negative: the liquid is depressed in the tube with a convex meniscus. Capillarity explains the rise of oil in a lamp wick, ink in blotting paper, and water through soil and in towels.

JEE Advanced

Tube shorter than $h$ : the liquid rises to the top and stops; it does not overflow. The meniscus flattens to a larger radius $R^{'}$ so that $h R = h^{'} R^{'}$ with $h^{'}$ the tube length (i.e. $cos θ^{'} = \frac{h ^{'}}{h} cos θ$ ). Freely falling lift or satellite: $g_{eff} = 0$ , so the liquid rises to the full length of the tube. In a lift accelerating up with $a$ , $h = \frac{2 T cos θ}{r ρ ( g + a )}$ .

Water in a glass tube

Adhesion $>$ cohesion, $θ < 9 0^{\circ}$ (about $0^{\circ}$ for clean glass). Concave meniscus; the liquid rises by $h \propto \frac{1}{r}$ .

Mercury in a glass tube

Cohesion $>$ adhesion, $θ \approx 14 0^{\circ}$ . Convex meniscus; $cos θ < 0$ , so the level is depressed.

Figure 10: Most surface tension slips come from miscounting surfaces. Film or bubble in air: two surfaces; drop or air bubble in a liquid: one.

Key idea

Surface tension is energy per unit area. Count surfaces: drops and air bubbles in a liquid have one (

\frac{2 T}{R}

), soap bubbles have two (

\frac{4 T}{R}

); capillary rise is

\frac{2 T cos θ}{r ρ g}

Quick Recall: tap to check

Excess pressure inside a soap bubble of radius

1 cm

(

T = 0.03 N m^{- 1}

\frac{4 T}{R} = \frac{0.12}{0.01} = 12 Pa

Two soap bubbles of different sizes are joined. Which way does air flow?

From the smaller bubble (higher excess pressure) into the larger one.

Water rises

6 cm

in a capillary. Rise in a tube of twice the radius?

3 cm

, since

h \propto \frac{1}{r}

3. Viscosity

When a liquid flows over a fixed surface, the layer touching the surface stays at rest and each higher layer moves a little faster. Adjacent layers rub against each other; this internal friction is viscosity. It opposes relative motion between layers, just as friction opposes sliding between solids.

Figure 11: Laminar flow between plates. Layer speed grows linearly from

0

v

; the force needed is

F = η A \frac{v}{l}

Newton's law of viscosity: the viscous force between two layers of area $A$ is proportional to the velocity gradient $\frac{d v}{d x}$ (change of speed per unit distance across the layers):

F = η A \frac{d v}{d x}

$η$ is the coefficient of viscosity: SI unit $N s m^{- 2} = Pa s$ (also called poiseuille, PI); CGS unit poise, with $1 Pa s = 10 poise$ ; dimensions $[ML^{- 1} T^{- 1}]$ .

Viscosity is also defined as the ratio of shearing stress to the rate of shear strain, $η = \frac{F / A}{v / l}$ , the fluid counterpart of the modulus of rigidity of a solid.

Fluid	Temperature ( $^{\circ} C$ )	$η$ (mPa s)
Water	20	1.0
Water	100	0.3
Blood	37	2.7
Machine oil	16	113
Glycerine	20	830
Honey	-	200
Air	40	0.019

The viscosity of liquids decreases as temperature rises (engine oil thins when hot); the viscosity of gases increases with temperature, because faster molecules carry more momentum between layers.

JEE Advanced

Poiseuille's formula. For steady laminar flow of a liquid of viscosity $η$ through a capillary of radius $r$ and length $L$ under a pressure difference $P$ , the speed at distance $y$ from the axis is

v = \frac{P}{4 η L} (r^{2} - y^{2})

a parabolic profile: fastest on the axis, zero at the wall. Adding up the flow through thin rings gives the volume flowing per second:

\frac{d Q}{d t} = \frac{π P r ^{4}}{8 η L}

The flow rate depends on $r^{4}$ : halving the radius of a tube (or a narrowed blood vessel) cuts the flow to one-sixteenth for the same pressure difference.

3.1 Stokes' law and terminal velocity

Stokes' law: a small sphere of radius $r$ moving slowly with speed $v$ through a fluid of viscosity $η$ feels a viscous drag

F = 6 π η r v

A sphere of density $ρ$ released in a liquid of density $σ$ accelerates at first. The drag grows with speed until the net force becomes zero; after that it falls at a constant terminal velocity $v_{t}$ .

Weight $= \frac{4}{3} π r^{3} ρ g$ (down). Buoyant force $= \frac{4}{3} π r^{3} σ g$ (up). Drag $= 6 π η r v$ (up).
At terminal velocity: $\frac{4}{3} π r^{3} (ρ - σ) g = 6 π η r v_{t}$ .
So
$v_{t} = \frac{2 r ^{2} ( ρ - σ ) g}{9 η}$

Figure 12: Drag

6 π η r v

grows until

m g = F_{B} + 6 π η r v_{t}

. Arrows drawn to scale at

v_{t}

for a ball

2.5

times as dense as the liquid:

F_{B} = 0.4 m g

, drag

= 0.6 m g

. The speed approaches

v_{t}

exponentially (curve plotted for linear drag).

$v_{t} \propto r^{2}$ : large raindrops fall faster than fine drizzle, and fog droplets hardly fall at all. If $σ > ρ$ , $v_{t}$ is negative and the body rises (air bubbles in a fizzy drink). Parachutes work by making the drag large so that the terminal velocity is small.

Exam Trick

Drops merge: $v_{t}$ scales as $n^{2/3}$ . If $n$ identical drops coalesce, $R = n^{1/3} r$ , so the new terminal velocity is $n^{2/3}$ times the old. Eight drops make one drop falling 4 times as fast; 27 drops make one falling 9 times as fast.

3.2 Streamline and turbulent flow; Reynolds number

Slow, smooth flow is laminar (streamline). As the speed rises past a critical velocity the flow becomes turbulent, with eddies and irregular mixing. Whether a flow is laminar or turbulent is decided by the dimensionless Reynolds number:

R_{e} = \frac{ρ v d}{η}

Here $d$ is the pipe diameter. $R_{e}$ compares inertial forces with viscous forces. In a pipe, flow is laminar for $R_{e} < 1000$ , turbulent for $R_{e} > 2000$ and unsteady in between. The critical velocity is $v_{c} = \frac{R _{e} η}{ρ d}$ : it is higher for viscous liquids and narrow pipes.

Figure 13:

R_{e} = \frac{ρ v d}{η}

. Flow in a pipe is laminar for

R_{e} < 1000

, unsteady for

1000

2000

and turbulent above about

2000

(NCERT values).

Key idea

Viscosity is fluid friction:

F = η A \frac{d v}{d x}

. A falling sphere reaches

v_{t} = \frac{2 r ^{2} ( ρ - σ ) g}{9 η}

when weight = buoyancy + Stokes drag.

Quick Recall: tap to check

Convert

η = 0.8 poise

to SI.

0.08 Pa s

(

1 Pa s = 10 poise

How does

v_{t}

change if the radius of the sphere doubles?

It becomes

4

times, since

v_{t} \propto r^{2}

Water at

5 cm s^{- 1}

in a

2 cm

pipe has

R_{e} = 1000

. Laminar or turbulent?

At the laminar limit: about 1000 is the upper end of laminar flow in a pipe.

Figure 14: Revision map: surface tension controls drops, bubbles and capillaries; viscosity controls drag, terminal velocity and the change from laminar to turbulent flow.

4. Solved Examples

Solved Example 1

What is the surface energy of a soap bubble of radius

r

Solution:

Surface energy $= T \times$ total surface area. A soap bubble has two surfaces (inner and outer), each of area $4 π r^{2}$ : $E = T \times 4 π r^{2} \times 2$ .

Answer: $E = 8 π r^{2} T$ .

Solved Example 2

A mercury drop of radius

1 cm

is sprayed into

1 0^{6}

droplets of equal size. Calculate the energy expended. (Surface tension of mercury

= 0.465 N m^{- 1}

)

Solution:

Energy expended = work done against surface tension $= T Δ S = T \cdot 4 π (n r^{2} - R^{2})$ .

Volume is conserved: $\frac{4}{3} π R^{3} = n \cdot \frac{4}{3} π r^{3}$ , so $r = \frac{R}{n ^{1/3}}$ and $n r^{2} = n^{1/3} R^{2}$ . Hence

E = 4 π R^{2} T (n^{1/3} - 1) = 4 π (1 0^{- 2})^{2} (0.465) (100 - 1)

Answer: $E \approx 5.8 \times 1 0^{- 2} J$ .

Solved Example 3

A metal plate of area

0.04 m^{2}

lies on a liquid layer of thickness

1 0^{- 3} m

and coefficient of viscosity

140

poise. Calculate the horizontal force needed to move the plate with a speed of

0.040 m s^{- 1}

Solution:

$η = 140 poise = 14 Pa s$ . The bottom layer is at rest, so the velocity gradient is $\frac{Δ v}{Δ x} = \frac{0.040}{1 0 ^{- 3}} = 40 s^{- 1}$ .

$F = η A \frac{Δ v}{Δ x} = 14 \times 0.04 \times 40$ .

Answer: $F = 22.4 N$ .

Solved Example 4

Find the work done in blowing a soap bubble of radius

5 cm

. Surface tension of soap solution

= 0.03 N m^{- 1}

Solution:

A bubble has two surfaces: $Δ A = 2 \times 4 π R^{2} = 8 π (0.05)^{2}$ .

$W = T Δ A = 0.03 \times 8 π \times 0.0025$ .

Answer: $W \approx 1.9 \times 1 0^{- 3} J$ .

Solved Example 5

One thousand water droplets, each of radius

1 mm

, merge into one drop. How much energy is released? (

T = 0.072 N m^{- 1}

)

Solution:

$R = n^{1/3} r = 10 \times 1 mm = 10 mm$ .

Energy released $= 4 π T (n r^{2} - R^{2}) = 4 π (0.072) [1000 (1 0^{- 3})^{2} - (1 0^{- 2})^{2}] = 4 π (0.072) (9 \times 1 0^{- 4})$ .

Answer: $\approx 8.1 \times 1 0^{- 4} J$ , released as heat.

Solved Example 6

Find the excess pressure inside a mercury drop of radius

3.00 mm

. (

T = 0.465 N m^{- 1}

)

Solution:

A drop has one surface: $Δ P = \frac{2 T}{R} = \frac{2 ( 0.465 )}{3 \times 1 0 ^{- 3}}$ .

Answer: $Δ P = 310 Pa$ .

Solved Example 7

An air bubble of radius

0.1 mm

10 cm

below the surface of water. Find the pressure inside it. (

P_{0} = 1.013 \times 1 0^{5} Pa

T = 0.073 N m^{- 1}

g = 9.8 m s^{- 2}

)

Solution:

Pressure in the water at that depth: $P_{0} + ρ g h = 1.013 \times 1 0^{5} + 980$ .

Excess inside (one surface): $\frac{2 T}{r} = \frac{2 ( 0.073 )}{1 0 ^{- 4}} = 1460 Pa$ .

Answer: $P = 1.013 \times 1 0^{5} + 980 + 1460 \approx 1.037 \times 1 0^{5} Pa$ .

Solved Example 8

How high does water rise in a glass capillary of radius

0.5 mm

? (

T = 0.073 N m^{- 1}

θ = 0^{\circ}

g = 9.8 m s^{- 2}

)

Solution:

$h = \frac{2 T cos θ}{r ρ g} = \frac{2 ( 0.073 ) ( 1 )}{( 5 \times 1 0 ^{- 4} ) ( 1000 ) ( 9.8 )}$ .

Answer: $h \approx 0.030 m = 3.0 cm$ .

Solved Example 9

A glass tube of radius

1 mm

is dipped in mercury. Find the depression of mercury in the tube. (

T = 0.465 N m^{- 1}

θ = 14 0^{\circ}

ρ = 13600 kg m^{- 3}

g = 9.8 m s^{- 2}

)

Solution:

$h = \frac{2 T cos θ}{r ρ g} = \frac{2 ( 0.465 ) ( cos 14 0 ^{\circ} )}{( 1 0 ^{- 3} ) ( 13 600 ) ( 9.8 )} = \frac{2 ( 0.465 ) ( - 0.766 )}{133.3}$ .

Answer: $h \approx - 5.3 mm$ , i.e. the mercury level is depressed by about $5.3 mm$ .

Solved Example 10

Water would rise

8 cm

in a capillary, but only

5 cm

of the tube sticks out above the water. What happens?
(A) Water overflows like a fountain
(B) Water rises to the top and stops, with a flatter meniscus
(C) Water does not rise at all
(D) Water rises to

5 cm

and the meniscus becomes more curved

Solution:

Answer: (B). The column stops at the rim. Since $h R$ is constant, the meniscus radius grows to $\frac{8}{5}$ of its original value (flatter), so the upward pull just supports $5 cm$ . Overflow would be a perpetual-motion machine.

Solved Example 11

A plate of area

0.05 m^{2}

slides at a steady

0.1 m s^{- 1}

over a

0.5 mm

thick layer of oil, pulled by a

0.02 kg

hanging mass through a string over a pulley. Find the viscosity of the oil. (

g = 9.8 m s^{- 2}

)

Solution:

Steady speed: viscous force = pull $= m g = 0.196 N$ .

$η = \frac{F / A}{v / l} = \frac{0.196/0.05}{0.1/ ( 5 \times 1 0 ^{- 4} )} = \frac{3.92}{200}$ .

Answer: $η \approx 1.96 \times 1 0^{- 2} Pa s$ .

Solved Example 12

A steel ball of radius

2 mm

falls through glycerine. Find its terminal velocity. (Density of steel

7800 kg m^{- 3}

, glycerine

1260 kg m^{- 3}

η = 0.83 Pa s

g = 9.8 m s^{- 2}

)

Solution:

$v_{t} = \frac{2 r ^{2} ( ρ - σ ) g}{9 η} = \frac{2 ( 2 \times 1 0 ^{- 3} ) ^{2} ( 7800 - 1260 ) ( 9.8 )}{9 ( 0.83 )} = \frac{0.513}{7.47}$ .

Answer: $v_{t} \approx 0.069 m s^{- 1} = 6.9 cm s^{- 1}$ .

Solved Example 13

Find the terminal velocity of a fog droplet of radius

0.01 mm

in air. (Viscosity of air

1.8 \times 1 0^{- 5} Pa s

, density of air

1.2 kg m^{- 3}

g = 9.8 m s^{- 2}

)

Solution:

$v_{t} = \frac{2 ( 1 0 ^{- 5} ) ^{2} ( 1000 - 1.2 ) ( 9.8 )}{9 ( 1.8 \times 1 0 ^{- 5} )}$ .

Answer: $v_{t} \approx 1.2 \times 1 0^{- 2} m s^{- 1}$ , about $1 cm s^{- 1}$ : fog drifts rather than falls.

Solved Example 14

Eight identical raindrops, each falling with terminal velocity

5 cm s^{- 1}

, coalesce into one drop. Find the new terminal velocity.
(A)

10 cm s^{- 1}

(B)

20 cm s^{- 1}

(C)

40 cm s^{- 1}

(D)

5 cm s^{- 1}

Solution:

Answer: (B). $R = 8^{1/3} r = 2 r$ and $v_{t} \propto r^{2}$ , so $v_{t}^{'} = 4 \times 5 = 20 cm s^{- 1}$ .

Solved Example 15

Water (

η = 1.0 \times 1 0^{- 3} Pa s

) flows at

5 cm s^{- 1}

through a pipe of diameter

2 cm

. Find the Reynolds number and the speed at which flow would turn turbulent (

R_{e} = 2000

Solution:

$R_{e} = \frac{ρ v d}{η} = \frac{( 1000 ) ( 0.05 ) ( 0.02 )}{1 0 ^{- 3}} = 1000$ : laminar.

$v_{c} = \frac{2000 η}{ρ d} = \frac{2000 ( 1 0 ^{- 3} )}{( 1000 ) ( 0.02 )}$ .

Answer: $R_{e} = 1000$ ; $v_{c} = 0.1 m s^{- 1}$ .

Solved Example 16

Two soap bubbles of radii

1 cm

and

2 cm

are joined by a tube with a tap. When the tap is opened (

T = 0.03 N m^{- 1}

),
(A) air flows from the

2 cm

bubble to the

1 cm

bubble
(B) air flows from the

1 cm

bubble to the

2 cm

bubble
(C) no air flows
(D) both bubbles shrink equally

Solution:

Answer: (B). Excess pressure $\frac{4 T}{R}$ : $\frac{4 ( 0.03 )}{0.01} = 12 Pa$ in the small bubble and $6 Pa$ in the large one. Air moves from high to low pressure, so the small bubble shrinks and the large one grows.

Solved Example 17

Water (

T = 0.072 N m^{- 1}

θ = 0^{\circ}

) rises in a clean glass capillary of radius

0.2 mm

. Find the upward force of surface tension on the column and the height of rise, and check that they agree. (

g = 9.8 m s^{- 2}

)

Solution:

Upward force $= 2 π r T cos θ = 2 π (2 \times 1 0^{- 4}) (0.072) = 9.05 \times 1 0^{- 5} N$ .

Height: $h = \frac{2 T cos θ}{r ρ g} = \frac{2 ( 0.072 )}{( 2 \times 1 0 ^{- 4} ) ( 1000 ) ( 9.8 )} = 0.0735 m$ .

Weight of the column $= π r^{2} h ρ g = π (2 \times 1 0^{- 4})^{2} (0.0735) (1000) (9.8) = 9.05 \times 1 0^{- 5} N$ , equal to the upward force.

Answer: $9.05 \times 1 0^{- 5} N$ ; $h \approx 7.3 cm$ .

Solved Example 18

For a liquid on a solid,

S_{s a} = 0.10 N m^{- 1}

S_{s l} = 0.05 N m^{- 1}

and

S_{l a} = 0.07 N m^{- 1}

. The angle of contact is about
(A)

0^{\circ}

(B)

4 4^{\circ}

(C)

9 0^{\circ}

(D)

13 6^{\circ}

Solution:

Answer: (B). $cos θ = \frac{S _{s a} - S _{s l}}{S _{l a}} = \frac{0.05}{0.07} = 0.714$ , so $θ \approx 4 4^{\circ}$ . Since $S_{s l} < S_{s a}$ the angle is acute and the liquid wets the solid.

Practice Questions

How much work is needed to increase the radius of a soap bubble from $1 cm$ to $2 cm$ ? ( $T = 0.03 N m^{- 1}$ )Answer: $8 π T (R_{2}^{2} - R_{1}^{2}) \approx 2.3 \times 1 0^{- 4} J$
Two soap bubbles have radii in the ratio $1 : 2$ . Find the ratio of their excess pressures.Answer: $2 : 1$
Water rises $3 cm$ in a capillary. How high will it rise in a tube of half the radius?Answer: $6 cm$
Find the capillary rise of water ( $T = 0.07 N m^{- 1}$ , $θ = 0$ ) in a tube of radius $0.1 mm$ . ( $g = 9.8 m s^{- 2}$ )Answer: $h \approx 0.14 m$
Two balls of the same material with radii $r$ and $2 r$ fall through the same oil. Compare their terminal velocities.Answer: $1 : 4$
Why does the viscosity of a gas increase while that of a liquid decreases on heating?Answer: Liquids: weaker cohesive forces when hot; gases: faster molecules transfer more momentum between layers

Common Mistakes to Avoid

Watch out

Using $\frac{2 T}{R}$ for a soap bubble. It has two surfaces: $\frac{4 T}{R}$ (and $8 π R^{2} T$ for its surface energy).
Using $\frac{4 T}{R}$ for an air bubble inside a liquid. It has only one surface: $\frac{2 T}{R}$ .
Forgetting $cos θ$ in Jurin's law, or using the diameter instead of the radius $r$ .
Thinking water overflows from a capillary that is shorter than $h$ . It rises to the top and the meniscus flattens.
Leaving out or reversing the buoyant term in terminal velocity: it is (density of body minus density of liquid), $(ρ - σ)$ .
Assuming viscosity always falls with temperature. That is true for liquids; for gases it rises.
Writing poise and pascal-second as equal: $1 Pa s = 10 poise$ .

Frequently Asked Questions

Which properties of fluids are important for NEET?

NEET regularly asks surface energy and work in blowing bubbles, excess pressure inside drops and soap bubbles, capillary rise with Jurin's law, the angle of contact, and terminal velocity from Stokes' law. Most are ratio questions, such as how the rise changes when the tube radius doubles.

What is surface tension and what is its SI unit?

Surface tension is the force per unit length acting along a liquid surface, or the energy needed per unit increase in surface area. Its SI unit is newton per metre, which is the same as joule per square metre.

Why are raindrops and soap bubbles spherical?

Surface tension makes a liquid surface behave like a stretched membrane that tries to reach the least possible area. For a given volume a sphere has the smallest surface area, so free drops and bubbles take a spherical shape.

Why is the excess pressure in a soap bubble 4T/R but in a drop only 2T/R?

A liquid drop has one surface, but a soap bubble is a thin film with an inner and an outer surface. Each surface contributes 2T/R, so the bubble's excess pressure is twice that of a drop of the same radius.

Why does water rise in a capillary tube but mercury falls?

Water wets glass because adhesion exceeds cohesion, giving an angle of contact below 90 degrees and a concave meniscus, so it rises. Mercury does not wet glass, its angle of contact is about 140 degrees, and its level is pushed down.

What is terminal velocity?

It is the constant speed reached by a body falling through a fluid when its weight is exactly balanced by the buoyant force and the viscous drag. For a small sphere it equals 2 r squared times the density difference times g, divided by 9 eta.

What surface tension and viscosity questions come in JEE Main and Advanced?

JEE Main asks energy released when drops merge, excess pressure in bubbles and capillary rise. JEE Advanced adds joined bubbles, capillaries shorter than the rise height, capillaries in accelerating lifts, charged bubbles, terminal velocity graphs and viscous force between plates.

What does the Reynolds number tell us?

The Reynolds number, rho v d over eta, compares inertial forces with viscous forces in a flow. In a pipe the flow is laminar when it is below about 1000 and turbulent when it is above about 2000.

Previous year questions on Properties of Fluids

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