The Surface Tension Formula gives the quantitative relationship between the force acting along a liquid surface and the length over which that force acts, expressed as \( T = F/l \). This fundamental concept appears in NCERT Class 11 Physics, Chapter 10 (Mechanical Properties of Fluids), and is a recurring topic in CBSE board exams as well as JEE Main and NEET. In this article, we cover the complete surface tension formula, its derivation, a comprehensive formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Surface Tension Formulas at a Glance
Quick reference for the most important surface tension formulas used in Class 11, CBSE, JEE, and NEET.
- Surface Tension: \( T = \dfrac{F}{l} \)
- Excess Pressure inside a liquid drop: \( \Delta P = \dfrac{2T}{r} \)
- Excess Pressure inside a soap bubble: \( \Delta P = \dfrac{4T}{r} \)
- Capillary Rise: \( h = \dfrac{2T\cos\theta}{r\rho g} \)
- Surface Energy: \( E = T \times A \)
- Work done to split a drop: \( W = 4\pi r^2 T \left(n^{1/3} – 1\right) \) (for n equal drops)
- Angle of Contact relation: \( T_{SA} = T_{SL} + T_{LA}\cos\theta \)
What is the Surface Tension Formula?
The Surface Tension Formula describes a property of liquid surfaces. It measures the force per unit length acting tangentially along the surface of a liquid. In simple terms, surface tension arises because liquid molecules at the surface experience a net inward pull from molecules below. This creates a thin, elastic-like film on the liquid surface.
According to NCERT Class 11 Physics, Chapter 10, surface tension is defined as the tangential force acting per unit length on an imaginary line drawn on the free surface of a liquid. The formula is:
\[ T = \frac{F}{l} \]
Here, \( T \) is surface tension, \( F \) is the force acting along the surface, and \( l \) is the length of the line over which the force acts. The SI unit of surface tension is Newton per metre (N/m), which is equivalent to Joule per square metre (J/m²). Surface tension explains everyday phenomena such as water droplets forming spheres, insects walking on water, and the rise of water in thin capillary tubes. It is a scalar quantity and depends on the temperature and nature of the liquid.
Surface Tension Formula — Expression and Variables
The primary Surface Tension Formula is written as:
\[ T = \frac{F}{l} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( T \) | Surface Tension | N/m (Newton per metre) |
| \( F \) | Force acting along the liquid surface | N (Newton) |
| \( l \) | Length of the line of contact | m (metre) |
| \( r \) | Radius of drop or bubble | m (metre) |
| \( \Delta P \) | Excess pressure inside drop/bubble | Pa (Pascal) |
| \( h \) | Height of capillary rise | m (metre) |
| \( \theta \) | Angle of contact | degree or radian |
| \( \rho \) | Density of liquid | kg/m³ |
| \( g \) | Acceleration due to gravity | m/s² |
| \( A \) | Surface area | m² |
| \( E \) | Surface energy | J (Joule) |
Derivation of the Surface Tension Formula
Consider a thin rectangular wire frame with a movable wire of length \( l \) on one side. A soap film is formed on this frame. The film has two surfaces, so the total length of contact is \( 2l \).
Step 1: Apply an external force \( F \) to pull the movable wire outward by a small distance \( dx \).
Step 2: The work done against surface tension is \( W = F \times dx \).
Step 3: The increase in surface area is \( \Delta A = 2l \times dx \) (two surfaces).
Step 4: Surface tension equals surface energy per unit area:
\[ T = \frac{W}{\Delta A} = \frac{F \times dx}{2l \times dx} = \frac{F}{2l} \]
For a single liquid surface (like a liquid drop), the formula simplifies to \( T = F/l \). This derivation confirms the dimensional formula of surface tension as \([MT^{-2}]\).
Complete Physics Formula Sheet: Surface Tension
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Surface Tension (basic) | \( T = F/l \) | F = force, l = length | N/m | Class 11, Ch 10 |
| Surface Energy | \( E = T \times \Delta A \) | T = surface tension, ΔA = change in area | J | Class 11, Ch 10 |
| Excess Pressure (liquid drop) | \( \Delta P = 2T/r \) | T = surface tension, r = radius | Pa | Class 11, Ch 10 |
| Excess Pressure (soap bubble) | \( \Delta P = 4T/r \) | T = surface tension, r = radius | Pa | Class 11, Ch 10 |
| Capillary Rise | \( h = 2T\cos\theta / (r\rho g) \) | T, θ = contact angle, r = tube radius, ρ = density, g = gravity | m | Class 11, Ch 10 |
| Work done to blow a soap bubble | \( W = 8\pi r^2 T \) | r = radius of bubble, T = surface tension | J | Class 11, Ch 10 |
| Work done to split a drop (n drops) | \( W = 4\pi r^2 T(n^{1/3} – 1) \) | r = original radius, n = number of drops | J | Class 11, Ch 10 |
| Young’s Equation (angle of contact) | \( T_{SA} = T_{SL} + T_{LA}\cos\theta \) | TSA, TSL, TLA = interfacial tensions | N/m | Class 11, Ch 10 |
| Radius of combined bubble | \( r = (r_1^3 + r_2^3)^{1/3} \) | r1, r2 = radii of original bubbles | m | Class 11, Ch 10 |
| Pressure inside combined bubble | \( P_1 r_1 + P_2 r_2 = P r \) | P = excess pressure, r = radius | Pa·m | Class 11, Ch 10 |
Surface Tension Formula — Solved Examples
Example 1 (Class 9-10 / Basic Level)
Problem: A horizontal wire of length 5 cm is placed on the surface of water. A force of 0.036 N is required to pull it off the surface. Calculate the surface tension of water.
Given:
- Length of wire, \( l = 5 \) cm \( = 0.05 \) m
- Force, \( F = 0.036 \) N
Step 1: Write the Surface Tension Formula: \( T = F / l \)
Step 2: Substitute values: \( T = 0.036 / 0.05 \)
Step 3: Calculate: \( T = 0.72 \) N/m
Answer
Surface tension of water = 0.72 N/m
Example 2 (Class 11-12 Level)
Problem: A soap bubble of radius 2 cm is formed using a soap solution whose surface tension is 0.03 N/m. Calculate (a) the excess pressure inside the bubble and (b) the work done in blowing the bubble from radius zero to 2 cm.
Given:
- Radius, \( r = 2 \) cm \( = 0.02 \) m
- Surface tension, \( T = 0.03 \) N/m
Part (a) — Excess Pressure:
Step 1: Use the formula for excess pressure inside a soap bubble (two surfaces): \( \Delta P = 4T/r \)
Step 2: Substitute: \( \Delta P = (4 \times 0.03) / 0.02 \)
Step 3: Calculate: \( \Delta P = 0.12 / 0.02 = 6 \) Pa
Part (b) — Work Done:
Step 1: Use \( W = 8\pi r^2 T \)
Step 2: Substitute: \( W = 8 \times 3.14159 \times (0.02)^2 \times 0.03 \)
Step 3: Calculate: \( W = 8 \times 3.14159 \times 0.0004 \times 0.03 \)
Step 4: \( W = 8 \times 3.14159 \times 0.000012 = 3.016 \times 10^{-4} \) J
Answer
(a) Excess pressure = 6 Pa
(b) Work done = 3.016 × 10²&sup4; J ≈ 3.02 × 10−4 J
Example 3 (JEE/NEET Level)
Problem: A large liquid drop of radius \( R \) is broken into 1000 identical smaller droplets. If the surface tension of the liquid is \( T \), find the change in surface energy. Also state whether energy is absorbed or released.
Given:
- Radius of original drop = \( R \)
- Number of smaller drops, \( n = 1000 \)
- Surface tension = \( T \)
Step 1: Find the radius of each small drop using volume conservation.
Volume of large drop = total volume of small drops:
\[ \frac{4}{3}\pi R^3 = 1000 \times \frac{4}{3}\pi r^3 \]
Step 2: Simplify: \( R^3 = 1000r^3 \Rightarrow r = R/10 \)
Step 3: Calculate initial surface area: \( A_i = 4\pi R^2 \)
Step 4: Calculate final surface area: \( A_f = 1000 \times 4\pi r^2 = 1000 \times 4\pi (R/10)^2 = 1000 \times 4\pi R^2/100 = 40\pi R^2 \times 1 = 40\pi R^2 \)
Let us be precise: \( A_f = 1000 \times 4\pi (R/10)^2 = 4000\pi R^2/100 = 40\pi R^2 \)
Step 5: Change in surface area: \( \Delta A = A_f – A_i = 40\pi R^2 – 4\pi R^2 = 36\pi R^2 \)
Step 6: Change in surface energy: \( \Delta E = T \times \Delta A = 36\pi R^2 T \)
Step 7: Since \( \Delta E > 0 \), energy is absorbed from the surroundings to break the large drop.
Answer
Change in surface energy = \( 36\pi R^2 T \). Energy is absorbed during the process.
CBSE Exam Tips 2025-26
- Memorise the two excess pressure formulas separately. A liquid drop has one surface, so \( \Delta P = 2T/r \). A soap bubble has two surfaces, so \( \Delta P = 4T/r \). Mixing these up is the single most common error in CBSE exams.
- Write the dimensional formula. Surface tension has dimensions \([MT^{-2}]\). CBSE frequently asks for dimensional analysis of surface tension in 2-mark questions.
- Link surface tension to surface energy. Remember that \( T = E/A \). This equivalence is tested in assertion-reason and MCQ questions.
- State the effect of temperature. Surface tension decreases with increasing temperature. It becomes zero at the critical temperature. This is a standard 1-mark question in CBSE 2025-26 pattern.
- We recommend practising the capillary rise formula \( h = 2T\cos\theta/(r\rho g) \) with numerical values. CBSE often gives 3-mark numericals on this formula.
- Use SI units consistently. Convert all lengths to metres and forces to Newtons before substituting into the Surface Tension Formula. Unit errors cost marks in board exams.
Common Mistakes to Avoid
- Confusing drop and bubble formulas: Students often apply \( \Delta P = 4T/r \) to a liquid drop. A liquid drop has only one free surface. Use \( \Delta P = 2T/r \) for drops and \( \Delta P = 4T/r \) for soap bubbles only.
- Forgetting the factor of 2 in soap film problems: A soap film has two surfaces. When calculating surface energy or work done, multiply the area by 2. Many students use only one surface area.
- Incorrect unit conversion: Surface tension is in N/m. If the length is given in cm or mm, always convert to metres first. Forgetting this step leads to answers that are off by a factor of 10 or 100.
- Misidentifying angle of contact: For water in glass, the angle of contact is acute (about 0°–8°), so \( \cos\theta \) is positive and the liquid rises. For mercury in glass, the angle is obtuse (about 135°), so \( \cos\theta \) is negative and the liquid is depressed. Swapping these leads to sign errors.
- Treating surface tension as a vector: Surface tension is a scalar quantity. It acts along the surface but has no directional component in the vector sense. Do not resolve it using vector addition rules.
JEE/NEET Application of Surface Tension Formula
In our experience, JEE aspirants encounter surface tension problems in at least one question per year in JEE Main. NEET also tests this topic under the Mechanical Properties of Fluids chapter. Here are the key application patterns to master.
Pattern 1: Splitting and Merging of Drops
JEE frequently asks about the energy change when a large drop is split into \( n \) smaller drops (or when drops merge). The key formula is:
\[ \Delta E = 4\pi r^2 T (n^{1/3} – 1) \]
where \( r \) is the radius of the original drop. When drops merge, energy is released. When a drop splits, energy is absorbed. This distinction is critical for JEE answer choices.
Pattern 2: Pressure Inside Bubbles and Drops
JEE Advanced problems often combine Bernoulli’s equation with excess pressure inside bubbles. You may need to find the velocity of air escaping from a punctured soap bubble. The excess pressure \( \Delta P = 4T/r \) acts as the driving pressure difference in Bernoulli’s equation. Refer to our article on Bernoulli’s Equation Formula for the combined approach.
Pattern 3: Capillary Rise and Contact Angle
NEET tests the capillary rise formula directly. JEE tests it indirectly through questions on the radius of the meniscus and the relationship between contact angle and wettability. Remember that for a tube of radius \( r \), the radius of the meniscus is \( r/\cos\theta \). This gives the modified capillary formula:
\[ h = \frac{2T}{r_{meniscus} \cdot \rho g} \]
Practice problems from NCERT Exemplar and previous JEE papers on this pattern. Our experts suggest solving at least 10 past JEE questions on surface tension before the exam.
FAQs on Surface Tension Formula
Explore More Physics Formulas
We hope this comprehensive guide on the Surface Tension Formula has strengthened your understanding of fluid mechanics. To build a complete foundation in Physics, explore these related resources on ncertbooks.net:
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For the official NCERT syllabus and textbook content, refer to the NCERT official website.