1. Physics
  2. Properties of Solids and Liquids
  3. Surface Tension and Capillarity

Properties of Solids and Liquids · JEE & NEET Physics

Surface Tension and Capillarity: notes and previous year questions

The stretchy skin on a liquid: γ as force per length and energy per area, pressure inside drops and bubbles, the angle of contact and capillary rise.

Surface Tension and Capillarity in short

  • Surface molecules are pulled inward, so a liquid surface acts like a stretched skin.
  • γ is force per length and energy per area; a film has two surfaces.
  • The pressure inside a drop is higher by 2γ/r; inside a soap bubble by 4γ/r.
  • Smaller drops and bubbles have more pressure inside.

1A stretchy skin

A molecule inside a liquid is pulled equally on every side; one at the surface has neighbours only below and beside it, so it is pulled inward. The surface therefore acts like a stretched skin that tries to shrink to the least area. That is why a water strider can stand on water, a needle can float, and small drops and bubbles are round (a sphere has the least surface for its volume).

2Force and energy

γ=FL=WΔA\gamma = \frac{F}{L} = \frac{W}{\Delta A}Force per unit length along a line in the surface, or work per unit area of new surface. N/m = J/m²; dimensions [M T⁻²].

A soap film has two surfaces, so on a sliding wire of length LL it pulls with F=2γLF = 2\gamma L: for γ=0.03\gamma = 0.03 N/m and L=10L = 10 cm, 6 mN.

Liquid (20 °C)γ (N/m)
Water0.073
Mercury0.465
Glycerin0.063
Soap solution≈ 0.025–0.03
Ethanol0.022

Surface tension falls as the liquid warms, and is zero at the critical temperature. Blowing a soap bubble of radius 5 cm (two surfaces): W=γ×8πr2≈1.9W = \gamma \times 8\pi r^2 \approx 1.9 mJ.

3Pressure inside drops and bubbles

Cut a drop in half: the excess pressure pushes on the flat face with ΔP πr2\Delta P\,\pi r^2, and surface tension pulls round the rim with γ 2πr\gamma\,2\pi r. Balancing them:

ΔP=2γr\Delta P = \frac{2\gamma}{r}A drop, or an air bubble inside a liquid (one surface).
ΔP=4γr\Delta P = \frac{4\gamma}{r}A soap bubble (two surfaces).

Smaller drops have more pressure inside: a 2 mm water drop 73 Pa, a 1 mm drop 146 Pa, a 1 µm droplet about 1.4 atm. Joined by a tube, a small bubble empties into a big one. A bubble grown from 3 cm to 4 cm (γ = 0.03 N/m) needs 1 Pa less excess pressure.

4Angle of contact

The angle of contact θ is measured through the liquid, between the solid and the liquid surface. It is a contest between adhesion (liquid–solid) and cohesion (liquid–liquid).

θWets?Meniscus
Adhesion wins< 90°yesconcave
Cohesion wins> 90°noconvex

Water on clean glass ≈ 0–10°; mercury on glass ≈ 140°; water on Teflon ≈ 108°; water on a lotus leaf ≈ 160° (it beads and rolls off, carrying dirt).

5Capillary rise

Surface tension pulls up round the inside rim (γcos⁡θ×2πr\gamma\cos\theta \times 2\pi r) until it balances the weight of the raised column (πr2hρg\pi r^2 h \rho g):

h=2γcos⁡θρgrh = \frac{2\gamma\cos\theta}{\rho g r}
  • h∝1/rh \propto 1/r: water rises 1.46 cm at r=1r = 1 mm, 2.92 cm at 0.5 mm, 5.84 cm at 0.25 mm. Twice the radius, half the rise.
  • Mercury (θ=140°\theta = 140°, cos⁡θ<0\cos\theta < 0) is pushed down: about 5.2 mm in a 1 mm tube.
  • A tilted tube reaches the same vertical height, so the column is longer (twice as long at 60° from the vertical).
  • A tube shorter than hh does not overflow: the surface just flattens.

6Surface tension at work

  • Soap lowers water's γ from 0.073 to about 0.025 N/m, so soapy water wets and soaks into greasy cloth.
  • Waterproof coatings make θ above 90°, so water beads and runs off.
  • Oil spreads on water: the water's stronger surface pulls it into a thin film.
  • In plants, xylem tubes (radius ≈ 0.01 mm) lift water about 1.5 m by capillarity; tall trees rely mostly on evaporation from the leaves pulling the water up.
  • Towels, wicks and sponges soak up liquids by capillarity.

Summary

Key ideas

  • Surface molecules are pulled inward, so a liquid surface acts like a stretched skin.
  • γ is force per length and energy per area; a film has two surfaces.
  • The pressure inside a drop is higher by 2γ/r; inside a soap bubble by 4γ/r.
  • Smaller drops and bubbles have more pressure inside.
  • The angle of contact decides whether a liquid wets a surface.
  • Capillary rise h = 2γ cos θ/(ρgr) grows as the tube gets thinner.
  • Liquids that do not wet (like mercury in glass) are pushed down instead.

Every equation

Surface tension
γ=F/L=W/ΔA\gamma = F/L = W/\Delta A
Soap film
F=2γLF = 2\gamma L
Bubble work
W=γ×8πr2W = \gamma\times 8\pi r^2
Drop
ΔP=2γ/r\Delta P = 2\gamma/r
Soap bubble
ΔP=4γ/r\Delta P = 4\gamma/r
Capillary rise
h=2γcos⁡θ/(ρgr)h = 2\gamma\cos\theta/(\rho g r)

Previous year questions with solutions

Real JEE and NEET questions on surface tension and capillarity. Try each one before you open the solution.

Q1JEE Main 2026One correct option

A liquid drop of diameter 2 mm breaks into 512 droplets. The change in surface energy is α×10−6\alpha \times {10}^{-6} J. The value of α\alpha is _______. (Take surface tension of liquid = 0.08 N/m)

  1. A10
  2. B7
  3. C8
  4. D11
Show answer and solution

Answer: Option B

The diameter is 2 mm2\ \mathrm{mm}, so R=1 mm=10−3 mR = 1\ \mathrm{mm} = 10^{-3}\ \mathrm{m}, and 5121/3=8512^{1/3} = 8:

ΔU=4πR2T (8−1)=4×3.14×10−6×0.08×7=7.04×10−6 J\Delta U = 4\pi R^{2}T\,(8 - 1) = 4 \times 3.14 \times 10^{-6} \times 0.08 \times 7 = 7.04 \times 10^{-6}\ \mathrm{J}, so α≈7\alpha \approx 7.

The trap is C, 88: it charges for the droplets' whole surface, 4πR2T×84\pi R^{2}T \times 8, forgetting that the original drop's surface was already there. The other trap is using the diameter as the radius, which would make the answer four times larger, 2828.

Q2NEET 2024One correct option

A thin flat circular disc of radius 4.5cm4.5 \mathrm{cm} is placed gently over the surface of water. If surface tension of water is 0.07Nm−10.07 Nm^{-1}, then the excess force required to take it away from the surface is

  1. A19.8 mN
  2. B198 N
  3. C1.98 mN
  4. D99 N
Show answer and solution

Answer: Option A

The water holds the disc along its rim only, a single circle of length 2πR2\pi R:

F=T×2πR=0.07×2×3.14×0.045=0.0198 N=19.8 mNF = T \times 2\pi R = 0.07 \times 2 \times 3.14 \times 0.045 = 0.0198\ \mathrm{N} = 19.8\ \mathrm{mN}.

The trap is C, 1.98 mN1.98\ \mathrm{mN}, a power of ten lost in converting 4.5 cm4.5\ \mathrm{cm} to 0.045 m0.045\ \mathrm{m}. B and D are forces of about a hundred newtons or more, the weight of a ten-kilogram mass, which no water surface around a small disc could supply. And the disc's area, πR2\pi R^{2}, plays no part: surface tension acts along a line.

Q3JEE Advanced 2020Numerical answer

When water is filled carefully in a glass, one can fill it to a height hh above the rim of the glass because of the surface tension of water. To calculate hh just before water starts flowing, model the water above the rim as a flat disc of thickness hh whose edge is rounded into a semicircle (seen in cross-section, the edge is a half-cylinder of diameter hh). When the pressure of water at the bottom of this disc exceeds what the curved surface at the edge can withstand, the surface breaks near the rim and water starts flowing. If the density of water is 103 kg m−310^3\ \mathrm{kg\,m^{-3}}, its surface tension 0.07 N m−10.07\ \mathrm{N\,m^{-1}} and g=10 m s−2g = 10\ \mathrm{m\,s^{-2}}, find hh in mm.

Show answer and solution

Answer: 3.74 mm

The edge is a half-cylinder of diameter hh, so its radius is h2\dfrac{h}{2} and the most excess pressure it can hold is that of a cylindrical surface:

Th/2=2Th\dfrac{T}{h/2} = \dfrac{2T}{h}

At the bottom of the disc the water is at ρgh\rho g h above atmospheric. The water breaks through when these are equal:

ρgh=2Th\rho g h = \dfrac{2T}{h}, so h=2Tρg=2×0.07103×10=1.4×10−5 m=3.74×10−3 m=3.74 mmh = \sqrt{\dfrac{2T}{\rho g}} = \sqrt{\dfrac{2 \times 0.07}{10^{3} \times 10}} = \sqrt{1.4 \times 10^{-5}}\ \mathrm{m} = 3.74 \times 10^{-3}\ \mathrm{m} = 3.74\ \mathrm{mm}

The trap is using the sphere's 2TR\dfrac{2T}{R}, which gives 4Tρg=5.29 mm\sqrt{\dfrac{4T}{\rho g}} = 5.29\ \mathrm{mm}: the edge is curved only in the vertical cross-section, not along the rim (the rim is so large that it is nearly straight). The other slip is taking the radius as hh instead of h2\dfrac{h}{2}, which gives 2.65 mm2.65\ \mathrm{mm}.

Practice questions, easy to hard

Three questions from the surface tension and capillarity practice ladder: one easy, one medium, one hard.

Q4One correct option

Where a liquid surface meets a solid, it pulls on the solid all along the line of contact, with TT for every metre of that line.

Lay a thin flat disc of radius RR on water and try to lift it straight up. The water surface clings to the rim and pulls it down, so beyond the disc's weight you need an extra

F=T×(length of the rim)=T×2πRF = T \times (\text{length of the rim}) = T \times 2\pi R

A thin wire ring is different: the water meets it along TWO circles, the ring's inner edge and its outer edge.

A thin wire ring and a thin flat disc, both of radius RR, rest on water and are lifted straight up. Compared with the disc, the extra force the ring needs is

  1. Athe same, since both rims have length 2πR2\pi R
  2. Bless, since the ring touches the water over a much smaller area
  3. Chalf as much
  4. Dtwice as much
Show answer and solution

Answer: Option D

The disc is held along one circle, its rim: 2πR T2\pi R\,T. The ring is held along its inner and its outer edge, and for a thin wire both circles have length very nearly 2πR2\pi R: 2×2πR T=4πR T2 \times 2\pi R\,T = 4\pi R\,T. Twice the disc's.

The trap is A, counting one circle for the ring as for the disc. B thinks in areas, but surface tension is a force per LENGTH: what matters is how long the line of contact is, not how much of the solid is wet. That is also why the disc's flat face adds nothing: the water under it pulls along no edge.

Q5One correct option

Pressures across surfaces add up as you go inward. A soap bubble of radius r1r_{1} blown inside a soap bubble of radius r2r_{2}: the air between them is at 4Tr2\dfrac{4T}{r_{2}} above the outside, and crossing the inner film adds another 4Tr1\dfrac{4T}{r_{1}}.

A soap bubble of radius 2 cm2\ \mathrm{cm} sits inside a soap bubble of radius 3 cm3\ \mathrm{cm}. The pressure in the inner bubble is the same as inside a single soap bubble of radius rr. What is rr?

  1. A5 cm5\ \mathrm{cm}
  2. B2 cm2\ \mathrm{cm}
  3. C1.2 cm1.2\ \mathrm{cm}
  4. D6 cm6\ \mathrm{cm}
Show answer and solution

Answer: Option C

Inner excess over the outside: 4T2+4T3=4Tr\dfrac{4T}{2} + \dfrac{4T}{3} = \dfrac{4T}{r}, so 1r=12+13=56\dfrac{1}{r} = \dfrac{1}{2} + \dfrac{1}{3} = \dfrac{5}{6} and r=1.2 cmr = 1.2\ \mathrm{cm}.

It must be smaller than 2 cm2\ \mathrm{cm}: a single bubble matching a pressure HIGHER than the inner bubble's own excess needs to be more sharply curved. The trap is A, 5 cm5\ \mathrm{cm}, adding the radii instead of their reciprocals. B ignores the outer film, and D is the product of the radii.

Q6One correct option

A soap film can hold up a load. On a U-shaped frame held upright with a horizontal slider at the bottom, the film pulls the slider up with 2TL2TL, and it stays at rest if 2TL2TL equals the weight hanging on it (slider included).

A light slider 6 cm6\ \mathrm{cm} long closes a vertical soap film (T=0.025 N/mT = 0.025\ \mathrm{N/m}). What mass hung from the slider keeps it at rest? (Take g=10 m/s2g = 10\ \mathrm{m/s^{2}}.)

  1. A0.15 g0.15\ \mathrm{g}
  2. B3 g3\ \mathrm{g}
  3. C0.3 g0.3\ \mathrm{g}
  4. D0.6 g0.6\ \mathrm{g}
Show answer and solution

Answer: Option C

mg=2TL=2×0.025×0.06=3×10−3 Nmg = 2TL = 2 \times 0.025 \times 0.06 = 3 \times 10^{-3}\ \mathrm{N}, so m=3×10−310=3×10−4 kg=0.3 gm = \dfrac{3 \times 10^{-3}}{10} = 3 \times 10^{-4}\ \mathrm{kg} = 0.3\ \mathrm{g}.

The trap is A, 0.15 g0.15\ \mathrm{g}, from one surface. B slips a power of ten converting kilograms to grams, and D counts the two surfaces twice over. A third of a gram is about right: a soap film can hold a small paper clip, not a coin.