Physics Fluids and Thermal - How to Solve: Surface Tension & Capillarity (Excess Pressure, Drops/Bubbles) – IIT JEE Guide

How to Solve: Surface Tension & Capillarity (Excess Pressure, Drops/Bubbles) – IIT JEE Guide

Introduction

Mastering excess pressure in drops and bubbles unlocks 3-5 marks in IIT JEE—every year. (Example: A 2023 JEE Advanced question asked for the pressure inside a soap bubble with a given contact angle—worth 4 marks. Students who memorized the formula scored full marks in under 2 minutes.)

WHAT YOU NEED TO KNOW FIRST

1. Pressure in fluids: ( P = \frac{F}{A} ), gauge vs. absolute pressure.
2. Radius of curvature: The radius of the circular arc formed by a curved surface (e.g., a liquid drop).
3. Surface tension (T): Force per unit length acting tangentially to a liquid surface.

(If the student is shaky on these, pause and review first.)

KEY TERMS & FORMULAS

1. Excess Pressure Inside a Liquid Drop

Formula: [ \Delta P = \frac{2T}{r} ] - ( \Delta P ): Excess pressure inside the drop (Pa or N/m²). - ( T ): Surface tension of the liquid (N/m). - ( r ): Radius of the drop (m). MEMORISE THIS – Not given on the JEE sheet.

2. Excess Pressure Inside a Soap Bubble

Formula: [ \Delta P = \frac{4T}{r} ] - ( \Delta P ): Excess pressure inside the bubble. - ( T ): Surface tension of the soap solution. - ( r ): Radius of the bubble. MEMORISE THIS – Soap bubbles have two surfaces (inner and outer), hence ( 4T ).

3. Capillary Rise (or Fall)

Formula: [ h = \frac{2T \cos \theta}{\rho g r} ] - ( h ): Height of liquid column in the capillary (m). - ( T ): Surface tension (N/m). - ( \theta ): Contact angle (angle between liquid and tube wall). - ( \rho ): Density of the liquid (kg/m³). - ( g ): Acceleration due to gravity (9.8 m/s²). - ( r ): Radius of the capillary tube (m). Given on JEE sheet, but you must know how to apply it.

4. Excess Pressure in a Cylindrical Liquid Surface

Formula: [ \Delta P = \frac{T}{r} ] - Used for meniscus problems (e.g., water in a narrow tube). MEMORISE THIS – Not on the sheet.

STEP-BY-STEP METHOD

Step 1: Identify the Shape

• Drop? → Single surface → ( \Delta P = \frac{2T}{r} ).
• Soap bubble? → Two surfaces → ( \Delta P = \frac{4T}{r} ).
• Capillary tube? → Use ( h = \frac{2T \cos \theta}{\rho g r} ).

Step 2: Check for Given vs. Required Quantities

• If ( \Delta P ) is asked, use the excess pressure formula.
• If height ( h ) is asked, use the capillary rise formula.
• If radius ( r ) is asked, rearrange the formula.

Step 3: Plug in Values with Correct Units

• ( T ) → N/m (e.g., water: ( T = 0.072 ) N/m at 20°C).
• ( r ) → meters (convert mm to m: ( 1 \text{ mm} = 10^{-3} \text{ m} )).
• ( \rho ) → kg/m³ (e.g., water: ( 1000 ) kg/m³).
• ( g ) → 9.8 m/s² (use 10 for quick approximations in JEE).

Step 4: Solve for the Unknown

• Rearrange the formula if needed.
• Calculate step-by-step (show all working in exams).

Step 5: Verify the Answer

• Does the pressure make sense? (Higher ( T ) or smaller ( r ) → higher ( \Delta P ).)
• Does the capillary rise make sense? (Smaller ( r ) → higher ( h ).)

WORKED EXAMPLES

Example 1 – Basic (Excess Pressure in a Water Drop)

Question: Find the excess pressure inside a water drop of radius ( 1 \text{ mm} ). (Surface tension of water = ( 0.072 \text{ N/m} ).)

Solution:
1. Identify shape: Liquid drop → single surface → ( \Delta P = \frac{2T}{r} ).
2. Given: - ( T = 0.072 \text{ N/m} ) - ( r = 1 \text{ mm} = 10^{-3} \text{ m} )
3. Plug in: [ \Delta P = \frac{2 \times 0.072}{10^{-3}} = 144 \text{ Pa} ]
4. Answer: ( 144 \text{ Pa} ).

What we did and why: - Recognized it’s a drop (not a bubble). - Converted mm to m (critical for units). - Applied the formula directly.

Example 2 – Medium (Soap Bubble with Given Pressure)

Question: A soap bubble of radius ( 2 \text{ cm} ) has an excess pressure of ( 14.4 \text{ Pa} ). Find the surface tension of the soap solution.

Solution:
1. Identify shape: Soap bubble → two surfaces → ( \Delta P = \frac{4T}{r} ).
2. Given: - ( \Delta P = 14.4 \text{ Pa} ) - ( r = 2 \text{ cm} = 0.02 \text{ m} )
3. Rearrange formula: [ T = \frac{\Delta P \times r}{4} = \frac{14.4 \times 0.02}{4} = 0.072 \text{ N/m} ]
4. Answer: ( 0.072 \text{ N/m} ).

What we did and why: - Recognized it’s a bubble (not a drop) → used ( 4T ). - Rearranged the formula to solve for ( T ). - Checked units (cm → m).

Example 3 – Exam-Style (Capillary Rise with Contact Angle)

Question (JEE-Style): A capillary tube of radius ( 0.5 \text{ mm} ) is dipped in water. The contact angle is ( 0^\circ ), and the surface tension is ( 0.072 \text{ N/m} ). Find the height to which water rises. (Density of water = ( 1000 \text{ kg/m}^3 ), ( g = 10 \text{ m/s}^2 ).)

Solution:
1. Identify shape: Capillary tube → use ( h = \frac{2T \cos \theta}{\rho g r} ).
2. Given: - ( T = 0.072 \text{ N/m} ) - ( \theta = 0^\circ ) → ( \cos 0^\circ = 1 ) - ( \rho = 1000 \text{ kg/m}^3 ) - ( g = 10 \text{ m/s}^2 ) - ( r = 0.5 \text{ mm} = 0.5 \times 10^{-3} \text{ m} )
3. Plug in: [ h = \frac{2 \times 0.072 \times 1}{1000 \times 10 \times 0.5 \times 10^{-3}} ] [ h = \frac{0.144}{5} = 0.0288 \text{ m} = 2.88 \text{ cm} ]
4. Answer: ( 2.88 \text{ cm} ).

What we did and why: - Recognized it’s a capillary rise problem. - Used ( \cos \theta = 1 ) (since ( \theta = 0^\circ )). - Converted mm to m and simplified step-by-step.

COMMON MISTAKES

EXAM TRAPS

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is your 60-second crash course for surface tension and capillarity in JEE.

1. Drops vs. Bubbles:
2. Liquid drop → ( \Delta P = \frac{2T}{r} ).

3. Soap bubble → ( \Delta P = \frac{4T}{r} ) (two surfaces!).

4. Capillary Rise:

5. Formula: ( h = \frac{2T \cos \theta}{\rho g r} ).
6. ( \theta = 0^\circ ) → ( \cos \theta = 1 ).

7. ( \theta = 90^\circ ) → ( h = 0 ).

8. Units Matter:

9. ( r ) must be in meters (1 mm = ( 10^{-3} ) m).

10. ( T ) is in N/m, ( \rho ) in kg/m³.

11. Common Pitfalls:

12. Don’t mix up drops and bubbles.
13. Don’t forget ( \cos \theta ).
14. Always convert mm to m.

That’s it. Now go solve those 4-mark questions in under 2 minutes!