Is it correct to say: Using 2T/R for a soap bubble?

A soap bubble in air has two surfaces, so ΔP = 4T/R and the work to blow it is 8πR²T. A liquid drop, or an air bubble inside a liquid, has one surface: 2T/R.

Is it correct to say: Substituting the diameter of the capillary tube for r?

h = 2T cos θ/(rρg) uses the radius of the bore. Halve the diameter and convert to metres first.

Is it correct to say: Ignoring buoyancy in terminal velocity?

Use (ρ − σ), the density of the sphere minus the density of the fluid. Only drop σ if the question says the fluid density is negligible (for example, air).

Is it correct to say: Mixing gauge pressure and absolute pressure?

P = P₀ + hρg is absolute pressure. hρg alone is gauge pressure. Read the question to see which one is asked.

Is it correct to say: Writing v ∝ 1/r in the equation of continuity?

A₁v₁ = A₂v₂ and A = πr², so v ∝ 1/r². Halving the radius makes the speed four times larger.

Is it correct to say: Expecting capillary rise for every liquid?

If the angle of contact is more than 90° (mercury in glass), cos θ is negative and the liquid level falls in the tube (capillary depression).

📄 Formula sheet (PDF)

Mechanical Properties of Fluids

Fluid pressure, Pascal's law, surface tension, capillarity, viscosity, Stokes' law, equation of continuity and Bernoulli's principle — Maharashtra HSC Physics Ch 2

📐 18 formulas✏️ 4 examples🎯 6 practice🏫 Maharashtra State Board📚 Class 12✓ 2025–26 syllabus

💡

Board Exam Tips

→The Maharashtra textbook writes surface tension as T (NCERT uses S). Use T in your answers so they match the textbook.
→A soap film or soap bubble has TWO surfaces; a liquid drop has ONE. Decide this first. It changes 2T/R to 4T/R and 4πR²T to 8πR²T.
→Standard derivations to practise: excess pressure inside a drop, capillary rise, terminal velocity from Stokes' law, the equation of continuity and Bernoulli's equation. Draw the diagram first; it makes the force or energy balance obvious.
→Convert mm and cm to metres before using h = 2T cos θ/(rρg) or ΔP = 2T/R. Radius, not diameter, goes into these formulas.
→For flow problems, use the equation of continuity first to get the second speed, then Bernoulli's equation for the pressure difference.

📐 Formulas(18)

Pressure

P = \frac{F}{A}

Symbol	Meaning
$P$	Pressure (Pa)
$F$	Normal force (N)
$A$	Area (m²)

Pressure due to a Liquid Column

P = h ρ g

Symbol	Meaning
$h$	Depth below the free surface (m)
$ρ$	Density of the liquid (kg m⁻³)
$g$	Acceleration due to gravity (m s⁻²)

Absolute and Gauge Pressure

P = P_{0} + h ρ g, P_{g a ug e} = P - P_{0} = h ρ g

Symbol	Meaning
$P$	Absolute pressure at depth h (Pa)
$P_{0}$	Atmospheric pressure (Pa)
$P_{g a ug e}$	Gauge pressure (Pa)

Pascal's Law — Hydraulic Lift

\frac{F _{1}}{A _{1}} = \frac{F _{2}}{A _{2}} \Rightarrow F_{2} = F_{1} \frac{A _{2}}{A _{1}}

Symbol	Meaning
$F_{1}, A_{1}$	Force (N) and area (m²) of the small piston
$F_{2}, A_{2}$	Force (N) and area (m²) of the large piston

Surface Tension

T = \frac{F}{l}

Symbol	Meaning
$T$	Surface tension (N m⁻¹)
$F$	Force due to the surface film (N)
$l$	Length of the line on which the force acts (m)

Surface Energy and Surface Tension★ Board fav

W = T Δ A \Rightarrow \frac{W}{Δ A} = T

Symbol	Meaning
$W$	Work done = increase in surface energy (J)
$Δ A$	Total increase in surface area (m²)

Work Done in Blowing a Soap Bubble

W = 8 π R^{2} T

Symbol	Meaning
$R$	Radius of the bubble (m)
$T$	Surface tension of the soap solution (N m⁻¹)

Splitting a Drop into n Identical Droplets

W = 4 π R^{2} T (n^{1/3} - 1)

Symbol	Meaning
$R$	Radius of the big drop (m)
$n$	Number of identical droplets formed

Excess Pressure inside a Liquid Drop (Laplace's Law)★ Board fav

P_{i} - P_{o} = \frac{2 T}{R}

Symbol	Meaning
$P_{i} - P_{o}$	Excess pressure, inside minus outside (Pa)
$T$	Surface tension (N m⁻¹)
$R$	Radius of the drop (m)

Excess Pressure inside a Soap Bubble

P_{i} - P_{o} = \frac{4 T}{R}

Symbol	Meaning
$R$	Radius of the soap bubble (m)

Capillary Rise★ Board fav

h = \frac{2 T c o s θ}{r ρ g}

Symbol	Meaning
$h$	Height of capillary rise (m)
$θ$	Angle of contact
$r$	Radius of the capillary bore (m)
$ρ$	Density of the liquid (kg m⁻³)

Newton's Law of Viscosity

F = η A \frac{d v}{d x}

Symbol	Meaning
$η$	Coefficient of viscosity (Pa s)
$A$	Area of the layer (m²)
$d v / d x$	Velocity gradient (s⁻¹)

Stokes' Law

F = 6 π η r v

Symbol	Meaning
$F$	Viscous force (N)
$r$	Radius of the sphere (m)
$v$	Speed of the sphere relative to the fluid (m s⁻¹)

Terminal Velocity★ Board fav

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

Symbol	Meaning
$v_{t}$	Terminal velocity (m s⁻¹)
$ρ$	Density of the material of the sphere (kg m⁻³)
$σ$	Density of the fluid (kg m⁻³)
$η$	Coefficient of viscosity of the fluid (Pa s)

Reynolds Number

R_{n} = \frac{v _{c} ρ d}{η}

Symbol	Meaning
$R_{n}$	Reynolds number (no unit)
$ρ$	Density of the fluid (kg m⁻³)
$v_{c}$	Critical velocity of the fluid (m s⁻¹)
$d$	Diameter of the pipe (m)

Equation of Continuity★ Board fav

A_{1} v_{1} = A_{2} v_{2}

Symbol	Meaning
$A_{1}, A_{2}$	Cross-sectional areas at two sections (m²)
$v_{1}, v_{2}$	Speeds of flow at those sections (m s⁻¹)

Bernoulli's Equation★ Board fav

P + \frac{1}{2} ρ v^{2} + ρ g h = constant

Symbol	Meaning
$P$	Pressure (Pa)
$ρ$	Density of the fluid (kg m⁻³)
$v$	Speed of flow (m s⁻¹)
$h$	Height above a reference level (m)

Speed of Efflux (Torricelli's Law)

v = 2 g h

Symbol	Meaning
$v$	Speed of efflux (m s⁻¹)
$h$	Depth of the hole below the free surface (m)

✏️ Solved Examples

1Solved Exampleeasy3 steps

Find the gauge pressure and the absolute pressure at a depth of 10 m below the surface of sea water. (Density of sea water = 1030 kg/m³, atmospheric pressure = 1.013 × 10⁵ Pa, g = 9.8 m/s²)

Gauge pressure is due to the liquid column alone.

P_{g a ug e} = h ρ g = 10 \times 1030 \times 9.8 = 1.009 \times 1 0^{5} Pa

2Solved Exampleboard3 steps

Water rises in a clean glass capillary tube of radius 0.2 mm. Find the height of rise. (Surface tension of water = 0.072 N/m, angle of contact = 0°, density = 1000 kg/m³, g = 9.8 m/s²)

Convert the radius to metres: r = 0.2 mm = 2 × 10⁻⁴ m. With θ = 0°, cos θ = 1.

h = \frac{2 T c o s θ}{r ρ g}

3Solved Exampleboard3 steps

A water drop of radius 1 cm is broken into 1000 identical droplets. Find the work done in the process. (Surface tension of water = 0.072 N/m)

Volume is conserved, so n^(1/3) = 1000^(1/3) = 10.

n^{1/3} = 10

4Solved ExampleHOTS3 steps

Water flows steadily through a horizontal pipe. At a wide section the area is 4 × 10⁻³ m² and the speed is 2 m/s. At a narrow section the area is 2 × 10⁻³ m². Find the speed at the narrow section and the pressure difference between the two sections. (Density of water = 1000 kg/m³)

Equation of continuity gives the speed at the narrow section.

v_{2} = \frac{A _{1} v _{1}}{A _{2}} = \frac{4 \times 1 0 ^{- 3} \times 2}{2 \times 1 0 ^{- 3}} = 4 m s^{- 1}

⚠️ Traps & Common Mistakes

⚠️Common Mistakes6

1
Using 2T/R for a soap bubble
✓A soap bubble in air has two surfaces, so ΔP = 4T/R and the work to blow it is 8πR²T. A liquid drop, or an air bubble inside a liquid, has one surface: 2T/R.
2
Substituting the diameter of the capillary tube for r
✓h = 2T cos θ/(rρg) uses the radius of the bore. Halve the diameter and convert to metres first.
3
Ignoring buoyancy in terminal velocity
✓Use (ρ − σ), the density of the sphere minus the density of the fluid. Only drop σ if the question says the fluid density is negligible (for example, air).
4
Mixing gauge pressure and absolute pressure
✓P = P₀ + hρg is absolute pressure. hρg alone is gauge pressure. Read the question to see which one is asked.
5
Writing v ∝ 1/r in the equation of continuity
✓A₁v₁ = A₂v₂ and A = πr², so v ∝ 1/r². Halving the radius makes the speed four times larger.
6
Expecting capillary rise for every liquid
✓If the angle of contact is more than 90° (mercury in glass), cos θ is negative and the liquid level falls in the tube (capillary depression).

🎯 Practice Yourself

🎯Practice Yourself6 questions

Q1
Find the excess pressure inside a soap bubble of radius 2 cm. (Surface tension of soap solution = 0.03 N/m)
Q2
In a hydraulic lift, the small piston has an area of 10 cm² and the large piston 500 cm². What load can be lifted by applying 100 N on the small piston?
Q3
A water droplet of radius 0.01 mm falls through air. Find its terminal velocity. (Viscosity of air = 1.8 × 10⁻⁵ Pa s, density of water = 1000 kg/m³, neglect the density of air, g = 9.8 m/s²)
Q4
Find the work done in blowing a soap bubble of radius 3 cm. (Surface tension of soap solution = 0.03 N/m)
Q5
Water flows at 3 m/s through a pipe of radius 2 cm, which narrows to a radius of 1 cm. Find the speed of water in the narrow part.
Q6
Find the speed of efflux of water from a small hole 5 m below the free surface of an open tank. (g = 9.8 m/s²)

📝 Notes

Mechanical Properties of Fluids — Maharashtra HSC Overview

Chapter 2 of the Maharashtra Board (Balbharati) Std XII Physics textbook deals with fluids at rest (pressure and surface tension) and fluids in motion (viscosity, the equation of continuity and Bernoulli's principle).

Fluids at rest

Pressure at depth $h$ is $P_0 + h\rho g$ . It depends only on depth, which explains the hydrostatic paradox and, through Pascal's law, the hydraulic lift ( $F_2 = F_1 A_2/A_1$ ).

Surface tension $T$ links three results that are worth learning as a chain:

Surface energy: work per unit increase in area equals $T$ .
Excess pressure: $2T/R$ for a drop, $4T/R$ for a soap bubble.
Capillary rise: $h = 2T\cos\theta/(r\rho g)$ . The liquid rises when $\theta < 90^\circ$ and is depressed when $\theta > 90^\circ$ .

Before any surface-tension numerical, decide how many free surfaces there are.

Fluids in motion

Viscosity is fluid friction between layers ( $F = \eta A\,dv/dx$ ). For a small sphere, Stokes' law $F = 6\pi\eta rv$ plus the weight and buoyancy gives the terminal velocity $v_t = 2r^2(\rho - \sigma)g/9\eta$ . The Reynolds number tells you whether the flow is streamline or turbulent.

For an ideal (non-viscous, incompressible) fluid in steady flow:

Continuity ( $A_1v_1 = A_2v_2$ ) is conservation of mass.
Bernoulli ( $P + \tfrac{1}{2}\rho v^2 + \rho gh$ = constant) is conservation of energy per unit volume.

Together they explain why the pressure is lower where a pipe narrows, why aerofoils experience lift, and Torricelli's speed of efflux $\sqrt{2gh}$ .

Approach for numericals

Write all lengths in metres, densities in kg m⁻³ and surface tension in N m⁻¹. In flow problems, find the unknown speed with continuity first, then apply Bernoulli between the same two points.