MHT-CET Physics · Mechanical Properties of Fluids

Surface Tension, Surface Energy, and Drops that Split or Merge

Surface tension is the force per unit length along a liquid surface, and equally the energy stored per unit area of surface; so stretching a film or a bubble costs T times the new area (both faces of a film), splitting a drop into many costs energy because the total surface grows, and merging drops releases it.

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Why this matters

42 PYQs, 12 HARD — the largest page in the chapter. Nineteen are drops splitting or merging (the work done, the final surface energy, the energy released and even the speed it gives the big drop); ten are work on films and bubbles, including the heated soap solution; six are forces along a line of contact — a coin, a paper disc with a hole, a drop leaving a tube, two plates squeezing a drop; seven are molecular and temperature facts. Four cards.

Concept 1 of 4: Surface Tension as a Force Along a Line

Surface tension T pulls along every line where the liquid surface meets a solid, with force T per metre of line. So count the lengths of contact: a coin's rim is 2πR; a paper disc with a hole has an outer rim 2πR and an inner rim 2πr, pulling together; a drop about to leave a tube hangs on the bore's circumference 2πr. A thin film between two glass plates is different: its curved edge makes a pressure deficit 2T/t across the whole area A, so the plates cling with F = 2TA/t, and with t = V/A that is 2TA²/V.

Definition

Force = $T \times$ (length of contact line).
Floating coin of radius R, thickness d, density ρ: $2\pi RT = \pi R^2 d\rho g \Rightarrow R = \dfrac{2T}{\rho g d}$ .
Disc with a hole: $2\pi T(R + r)$ — both rims.
Drop leaving a tube of bore radius r (zero contact angle): weight $w = 2\pi rT$ .
Drop squeezed between plates into a film of area A, thickness $t = V/A$ : $F = \dfrac{2TA}{t} = \dfrac{2TA^2}{V}$ , so $T = \dfrac{FV}{2A^2}$ .
Drop floating half immersed (density ρ in liquid d): weight = upthrust + $2\pi rT$ gives diameter $\sqrt{\dfrac{12T}{g(2\rho - d)}}$ .

Film between two plates

F = \frac{2TA}{t} = \frac{2TA^2}{V}

Worked example

0.02 cm³ of water (T = 0.07 N/m) is squeezed between glass plates into 20 cm². Force to pull them apart?

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The same idea in a real exam question:

MHT-CET · 2025 · 20 April Shift I · Q42Hard

Example 1 · Mechanical Properties of Fluids · Surface Tension and Surface Energy

A water drop of

0.01\,\text{cm}^3

is squeezed between two glass plates and spreads into an area of

10\,\text{cm}^2

. If surface tension of water is 70 dyne/cm, then the normal force required to separate the glass plates from each other will be

Counting only the outer rim of a ring

A disc with a hole touches the liquid along two circles. Both pull: 2πT(R + r), not 2πT(R − r).

Drill 5 more on surface tension as a force along a line

Concept 2 of 4: Work on Films and Soap Bubbles

Surface energy per unit area is T, so work = T × (new surface). A soap film and a soap bubble have TWO surfaces, front and back — double the area. A bubble of radius R therefore holds 8πR²T; growing it from diameter d to D costs 2π(D² − d²)T. Warming the solution lowers T, so a bubble twice as wide needs a little less than four times the work. And since area goes as volume to the 2/3, doubling a bubble's volume multiplies the work by 4^(1/3).

Definition

Film (two faces): $W = 2T\,\Delta A$ , where $\Delta A$ is the growth of ONE face. Two wires of length l moved apart by x: $W = 2Tlx$ .
Soap bubble (two faces): $W = 8\pi R^2 T$ ; diameter d → D: $W = 2\pi(D^2 - d^2)T$ .
Heated solution: T falls, so radius 2R needs slightly less than $4W$ .
Volume doubled: $W \propto R^2 \propto V^{2/3}$ , so $W' = 4^{1/3}W$ .

Soap bubble

W = 2 \times 4\pi R^2\,T = 8\pi R^2 T

Worked example

A soap film on a frame grows from 4 cm × 4 cm to 6 cm × 6 cm; T = 0.03 N/m. Work done?

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The same idea in a real exam question:

MHT-CET · 2025 · 25 April Shift II · Q29Moderate

Example 2 · Mechanical Properties of Fluids · Surface Tension and Surface Energy

The amount of work done in blowing a soap bubble such that its diameter increases from 'd' to '

D

' is (

T=

surface tension of solution)

Forgetting the second surface

A soap film or bubble has two faces. A liquid DROP has one. Using 4πR²T for a bubble halves the answer.

Drill 9 more on work on films and soap bubbles

Concept 3 of 4: Drops that Split or Merge

Volume is conserved, surface is not. Split one drop of radius R into n equal droplets: each has radius R/n^(1/3), and the total surface grows n^(1/3) times. So the work done is 4πR²T(n^(1/3) − 1), and the final surface energy is n^(1/3) times the first — 512 droplets give 8E, 729 give 9E, 1000 give 10E. Run it backwards and n droplets merging release energy E(n − n^(2/3)) where E is one droplet's surface energy. If that released energy all becomes kinetic energy of the big drop, v = √[(6T/ρ)(1/r − 1/R)].

Definition

Radius: n droplets of radius r make one of $R = n^{1/3}r$ (general: $R^3 = R_1^3 + R_2^3 + \dots$ ).
Split R into n: $W = 4\pi R^2T\left(n^{1/3} - 1\right) = 4\pi TR^2\left(\dfrac{R}{r} - 1\right)$ . n = 8 → $4\pi R^2T$ ; 64 → $12\pi R^2T$ ; 1000 → $36\pi R^2T$ .
Surface energy ratio (droplets : big) $= n^{1/3} : 1$ ; final : initial for merging 1000 = 1 : 10; eight mercury drops before : after = 2 : 1.
Merge n droplets each of energy E: released $E\left(n - n^{2/3}\right)$ .
Energy lost = 3E (E the big drop's): $n = \dfrac{4R^2}{r^2}$ .
Speed of the big drop (all released energy → KE): $v = \sqrt{\dfrac{6T}{\rho}\left(\dfrac{1}{r} - \dfrac{1}{R}\right)}$ .

Splitting a drop

W = 4\pi R^2 T\left(n^{1/3} - 1\right), \qquad \frac{E_{\text{droplets}}}{E_{\text{drop}}} = n^{1/3}

Worked example

A drop of radius R splits into 27 equal droplets. Work done, in terms of T and R?

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The same idea in a real exam question:

MHT-CET · 2024 · 11th May Shift 2 · Q5Hard

Example 3 · Mechanical Properties of Fluids · Surface Tension and Surface Energy

A liquid drop of radius 'R' is broken into 'n' identical small droplets. The work done is [

T =

surface tension of the liquid]

Using n^(2/3) for the total surface

Each droplet's area scales as n^(−2/3), and there are n of them: n × n^(−2/3) = n^(1/3). The factor n^(2/3) is the surface of the MERGED drop measured in droplet units, which is where E(n − n^(2/3)) comes from.

Drill 18 more on drops that split or merge

Concept 4 of 4: Surface Molecules, Temperature and Impurities

A molecule on the surface has neighbours on one side only, so it is pulled inward and has MORE potential energy than one inside; bringing molecules to the surface costs energy, which is why the surface tends to shrink. Heating weakens the attraction, so surface tension falls with temperature and vanishes at the critical temperature. A soluble impurity like soap lowers surface tension — easier spraying, smaller contact angle.

Definition

Surface molecules have maximum (larger) potential energy than molecules inside.
Temperature up → surface tension down; at the critical temperature it is zero.
Soap or detergent lowers surface tension, so water sprays more easily; a soluble impurity decreases the angle of contact.

Change	Surface tension	Angle of contact
Temperature rises	decreases	—
Critical temperature	zero	—
Soap / soluble impurity added	decreases	decreases A highly soluble salt can raise T slightly; the paper's answer is 'decreases' for soap and detergents.

Surface energy per unit area = surface tension.

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The same idea in a real exam question:

MHT-CET · 2024 · 14th May Shift 1 · Q6Easy

Example 4 · Mechanical Properties of Fluids · Surface Tension and Surface Energy

The potential energy of a molecule on the surface of a liquid compared to the molecules inside the liquid is

Thinking surface molecules have LESS energy

They have fewer neighbours, so fewer attractive bonds holding them down — their potential energy is higher, not lower.

Drill 6 more on surface molecules, temperature and impurities

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

Surface Tension as a Force Along a Line
Film between two plates
$F = \frac{2TA}{t} = \frac{2TA^2}{V}$
Work on Films and Soap Bubbles
Soap bubble
$W = 2 \times 4\pi R^2\,T = 8\pi R^2 T$
Drops that Split or Merge
Splitting a drop
$W = 4\pi R^2 T\left(n^{1/3} - 1\right), \qquad \frac{E_{\text{droplets}}}{E_{\text{drop}}} = n^{1/3}$

Reference tables (1)

Surface Molecules, Temperature and Impurities3 rows

Change	Surface tension	Angle of contact
Temperature rises	decreases	—
Critical temperature	zero	—
Soap / soluble impurity added	decreases	decreases A highly soluble salt can raise T slightly; the paper's answer is 'decreases' for soap and detergents.

Surface energy per unit area = surface tension.

Watch out for (4)

Counting only the outer rim of a ring
→ Surface Tension as a Force Along a Line
Forgetting the second surface
→ Work on Films and Soap Bubbles
Using n^(2/3) for the total surface
→ Drops that Split or Merge
Thinking surface molecules have LESS energy
→ Surface Molecules, Temperature and Impurities

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