Definition
Surface tension
Force per unit length acting tangentially on a liquid surface, opposing increase in surface area. T = F/L (units: N/m). Surface energy = T × ΔA when surface area increases by ΔA.
-- NCERT Class 11 Physics, Ch. 9, p. 194Surface Energy Tension
8 MCQs2 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsPYQ coverage: NEET 2023, 2024, 2025Official key: NTA-verifiedLast updated: 25 Sep 2026
Surface Energy Tension, explained for NEET
The high-frequency trap in surface energy and tension questions is confusing the number of free surfaces. A soap bubble has two surfaces (inner and outer); a liquid drop in air has one. This single distinction is the difference between choosing 4T/r and 2T/r for excess pressure — and between the correct and a common wrong option on exam day.
Surface tension is the force per unit length acting along the surface of a liquid, perpendicular to any line drawn on the surface and tangential to the surface itself (NCERT Class 11 Physics, Chapter 9, page 191). Its SI unit is N/m. The underlying cause is the net inward pull on surface molecules — they have fewer neighbours above than below, so the surface layer behaves like a stretched elastic membrane.
Surface energy is the work done per unit area to increase the free surface. For a liquid with surface tension T, creating a new surface of area ΔA costs work W = T × ΔA. This connects directly to the excess-pressure formula: for a spherical liquid drop (one free surface), ΔP = 2T/r; for a soap bubble in air (two free surfaces), ΔP = 4T/r (NCERT Class 11 Physics, Chapter 9, page 193).
Where NEET tests this: questions ask you to calculate the work done in forming a bubble, or to find excess pressure inside a bubble versus a drop. The trap is mechanical — you read "bubble," your hand writes 2T/r because the drop formula is more rehearsed. The fix: count surfaces first, write the formula second.
Watch out: an air bubble inside a liquid also has only one surface (liquid-air interface on the inside). It follows the drop formula, 2T/r, not the bubble formula.
Can you answer these Surface Energy Tension MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Surface tension of a liquid is defined as the force per unit length acting on a line drawn on the liquid surface. Its SI unit is:
Show answer and why every option is right or wrong
Answer: B. Surface tension is force per unit length, so its unit is N/m (NCERT Class 11 Physics, Chapter 9, page 191). While N/m is dimensionally equivalent to J/m², the standard SI unit for surface tension is expressed as N/m.
Why A is wrong: A (J/m²) is dimensionally equivalent to N/m but is the unit conventionally used for surface energy per unit area, not how surface tension is defined. NEET options distinguish the two.
Why C is wrong: C (N/m²) is the unit of pressure (pascal), not surface tension. Confusing force/length with force/area is a common unit-mismatch error.
Why D is wrong: D (J/m) has dimensions of force, not force per length. This does not match the definition.
MCQ 2Easy RecallPractice
Surface molecules of a liquid experience a net inward force because:
Show answer and why every option is right or wrong
Answer: C. Molecules at the surface have liquid neighbours only on the side facing the bulk, and fewer (or no liquid) neighbours above. The resulting asymmetric cohesive attraction produces a net inward pull (NCERT Class 11 Physics, Chapter 9, page 191).
Why A is wrong: A is wrong. Gravity acts on all molecules equally throughout the liquid; the net inward pull on surface molecules is due to asymmetric cohesive forces, not gravity.
Why B is wrong: B is wrong. Atmospheric pressure acts on the surface but does not explain the tangential contractile tendency of the surface. The net inward pull is a molecular cohesion effect, not a pressure effect.
Why D is wrong: D is wrong. Ordinary liquids do not carry a net surface charge. Surface tension is a cohesive-force phenomenon, not an electrostatic one.
MCQ 3Easy RecallPractice
The work done to increase the surface area of a liquid by ΔA, if the surface tension is T, equals:
Show answer and why every option is right or wrong
Answer: B. Surface energy equals surface tension multiplied by the change in surface area: W = T × ΔA (NCERT Class 11 Physics, Chapter 9, page 193).
Why A is wrong: A is wrong. Dividing T by ΔA gives units of N/m³, which has no physical meaning in this context. Work = T × ΔA, not T / ΔA.
Why C is wrong: C is wrong. Squaring the area gives units of N·m³, not joules. The relationship is linear in ΔA.
Why D is wrong: D is wrong. Squaring T gives units of N²/m², which is not an energy unit. The formula is first power in both T and ΔA.
MCQ 4Direct ApplicationPractice
A soap bubble of radius 1.0 × 10⁻² m is blown in air. The surface tension of the soap solution is 3.0 × 10⁻² N/m. The excess pressure inside the bubble over atmospheric pressure is:
Show answer and why every option is right or wrong
Answer: A. A soap bubble in air has two free surfaces. Excess pressure ΔP = 4T/r = 4 × 3.0 × 10⁻² / 1.0 × 10⁻² = 12 Pa (NCERT Class 11 Physics, Chapter 9, page 193).
Why B is wrong: B (6.0 Pa) results from using ΔP = 2T/r, which applies to a liquid drop (one surface), not a soap bubble (two surfaces). This is the bubble-vs-drop trap.
Why C is wrong: C (3.0 Pa) results from using ΔP = T/r, omitting the numerical factor entirely. Neither the drop nor the bubble formula gives T/r.
Why D is wrong: D (24 Pa) results from using ΔP = 8T/r, doubling the correct factor. There is no configuration with a factor of 8.
MCQ 5Direct ApplicationPractice
An air bubble of radius r sits inside a liquid of surface tension T. The excess pressure inside the air bubble compared to the surrounding liquid is:
Show answer and why every option is right or wrong
Answer: A. An air bubble inside a liquid has only one free surface (the liquid-air interface on the inner side). Therefore ΔP = 2T/r, identical to a liquid drop in air (NCERT Class 11 Physics, Chapter 9, page 193).
Why B is wrong: B (T/r) has no standard physical basis; the minimum factor for one surface is 2.
Why C is wrong: C (4T/r) applies to a soap bubble in air, which has two surfaces. An air bubble in liquid has only one surface.
Why D is wrong: D (8T/r) has no physical basis — no standard configuration yields a factor of 8.
MCQ 6Direct ApplicationPractice
The work done in blowing a soap bubble of radius R from soap solution of surface tension T is:
Show answer and why every option is right or wrong
Answer: D. A soap bubble has two surfaces (inner and outer). Total new surface area = 2 × 4πR² = 8πR². Work = T × total area = 8πR²T.
Why A is wrong: A (4πR²T) counts only one surface of the bubble. A soap bubble in air has an inner and outer surface, so the total area created is 8πR², not 4πR².
Why B is wrong: B (πR²T) confuses the cross-sectional area πR² with the spherical surface area 4πR². Surface energy requires the full surface area.
Why C is wrong: C (2πR²T) uses half the area of even a single spherical surface. There is no physical justification for the factor 2πR².
MCQ 7CalculationPYQ Pattern
Two soap bubbles of radii r₁ and r₂ (r₁ < r₂) are connected by a tube. What happens?
Show answer and why every option is right or wrong
Answer: D. Excess pressure ΔP = 4T/r. The smaller bubble (smaller r) has higher internal pressure. Air flows from higher pressure to lower pressure — from the smaller bubble to the larger one. The smaller bubble shrinks and the larger one grows.
Why A is wrong: A is wrong. The smaller bubble has higher excess pressure (4T/r with smaller r gives larger ΔP), so air flows out of it, not into it.
Why B is wrong: B is wrong. There is no mechanism for simultaneous collapse; pressure difference drives one-directional flow from the smaller to the larger bubble.
Why C is wrong: C is wrong. Equilibrium would require equal pressures, which demands equal radii. Since r₁ ≠ r₂, pressures differ and air flows.
MCQ 8CalculationPYQ Pattern
A soap bubble of radius 2.0 × 10⁻² m shrinks to radius 1.0 × 10⁻² m. The surface tension of the soap solution is 2.5 × 10⁻² N/m. The energy released during the shrinkage is closest to:
Show answer and why every option is right or wrong
Answer: C. A soap bubble has two surfaces, so total surface area = 8πR². Initial total area = 8π(2.0 × 10⁻²)² = 32π × 10⁻⁴ m². Final total area = 8π(1.0 × 10⁻²)² = 8π × 10⁻⁴ m². ΔA = 24π × 10⁻⁴ m². Energy released = T × ΔA = 2.5 × 10⁻² × 24π × 10⁻⁴ = 60π × 10⁻⁶ ≈ 1.88 × 10⁻⁴ J ≈ 1.9 × 10⁻⁴ J.
Why A is wrong: A (2.5 × 10⁻⁴ J) uses only the initial area, T × 8π(2.0 × 10⁻²)², instead of the change in area; the bubble still has surface at the smaller radius.
Why B is wrong: B (9.4 × 10⁻⁵ J) is approximately half the correct answer, resulting from counting only one surface of the soap bubble (4πR² instead of 8πR²).
Why D is wrong: D (3.8 × 10⁻⁴ J) is exactly double the correct answer. This could arise from incorrectly computing the area difference as 48π × 10⁻⁴ instead of 24π × 10⁻⁴.
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Surface Energy Tension: quick recall before you leave
How do you solve a Surface Energy Tension question? A worked example
- 1
Given
A spherical mercury drop of radius R = 1.0 × 10⁻³ m breaks up into 1000 identical smaller droplets. The surface tension of mercury is T = 0.465 N/m, and the temperature does not change.
- 2
Required
The work that has to be supplied to bring this about.
- 3
Concept
Surface energy is the work needed per unit area of NEW free surface: W = T × ΔA (NCERT Class 11 Physics, Chapter 9, page 194). Nothing else about the liquid changes here, so the whole energy cost is the extra area.
Two facts do the work. First, volume is conserved, which fixes the small radius. Second, a drop has ONE free surface — unlike a soap bubble, which has two — so each sphere contributes 4πr², not 8πr². Dividing one drop into many is the standard way of manufacturing a large area increase from a fixed volume: the volume goes as r³ while the area goes as r², so cutting the radius by ten multiplies the total area by ten. - 4
Formula
• Volume conservation: (4/3)πR³ = n × (4/3)πr³, so r = R / n^(1/3)• Area change: ΔA = n × 4πr² − 4πR²• Work: W = T × ΔA
- 5
Substitution
n^(1/3) = 1000^(1/3) = 10, so r = 1.0 × 10⁻³ / 10 = 1.0 × 10⁻⁴ m
Initial area = 4π × (1.0 × 10⁻³)²
Final area = 1000 × 4π × (1.0 × 10⁻⁴)²
W = 0.465 × (final area − initial area) - 6
Calculation
Initial area = 4π × 1.0 × 10⁻⁶ = 1.257 × 10⁻⁵ m²
Final area = 1000 × 4π × 1.0 × 10⁻⁸ = 1000 × 1.257 × 10⁻⁷ = 1.257 × 10⁻⁴ m²
ΔA = 1.257 × 10⁻⁴ − 1.257 × 10⁻⁵ = 1.131 × 10⁻⁴ m²
W = 0.465 × 1.131 × 10⁻⁴ = 5.26 × 10⁻⁵ J ≈ 5.3 × 10⁻⁵ J
Cross-check with the compact form W = 4πR²T(n^(1/3) − 1): 1.257 × 10⁻⁵ × 0.465 × (10 − 1) = 5.26 × 10⁻⁵ J. ✓
Note on exact values: n = 1000 is a count, and 4π and the exponents are mathematical constants. The two significant figures in R set the precision. - 7
Final answer
W ≈ 5.3 × 10⁻⁵ J must be supplied.
The area went up by a factor of ten while the volume stayed put — that factor is n^(1/3), and it is where all the energy went. Left alone the droplets coalesce again, releasing this energy as heat, which is why a fine mercury spray warms slightly as it recombines. - 8
Common trap
The bubble formula gets used here by reflex: 2 × 4πr² per sphere, doubling the answer to 1.05 × 10⁻⁴ J. A DROP has one surface. Count the surfaces before counting anything else — one for a drop and for an air bubble inside a liquid, two for a soap bubble in air.
The second trap is computing r as R/1000 instead of R/1000^(1/3). That is a thousand-fold error in the radius and a millionfold one in the area. - 9
Similar NEET-style question
Eight identical droplets of mercury, each of radius r, coalesce into one large drop. Is energy absorbed or released, and how much? Approach: the same relation run backwards. R = 8^(1/3) r = 2r, so the area falls from 8 × 4πr² to 4π(2r)² = 16πr² — exactly half. Energy is RELEASED, and it equals T × 16πr².
What to remember before solving Surface Energy Tension questions
Liquid drop: ΔP = 2T/r (one surface). Soap bubble: ΔP = 4T/r (two surfaces). Smaller drops have higher internal pressure.
-- NCERT Class 11 Physics, Ch. 9, p. 196More in Properties of Bulk Matter: 5 exam traps and mistakes · 12 formulas · 4 question patterns from its other lessons.
Surface Energy Tension questions from past NEET papers
3 questions from NEET 2023, 2024, 2025. Answers verified against NTA official keys. — click to collapse
NEET 2025
1a = 1/2, α = 1/2, β = −1/2, γ = 1/2, δ = 7/2
2a = 1/2, α = 1/2, β = −1, γ = +1, δ = 3/2
3a = −1/2, α = −1/2, β = −1, γ = −1/2, δ = 5/2
4a = −1/2, α = −1/2, β = −1, γ = 1/2, δ = 7/2
NTA Answer: Option 4(final)
NEET 2024Revised key
119.8 mN
2198 N
31.98 mN
499 N
NTA Answer: Option 1(revised_final)
NEET 2023
15.06 × 10⁻⁴ J
23.01 × 10⁻⁴ J
350.1 × 10⁻⁴ J
430.16 × 10⁻⁴ J
NTA Answer: Option 2(final)
All 17 past-paper questions from Properties of Bulk Matter →
How does NEET ask about Surface Energy Tension?
1 recurring pattern from past papers — click to collapse
Energy to form a bubble of given radius from soap solution; or pressure inside vs outside. Bubble has 2 surfaces (factor 4T/r).
Direct ApplicationEasy
Common distractors
uses 2T r instead of 4T r for bubble
Treats soap bubble like a drop
Sources
NCERT refs: Class 11 Physics Chapter 9, p.191 | Class 11 Physics Chapter 9, p.193
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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