Welcome to Lecture 87 of our comprehensive JEE Physics Master Series. In this lecture, we investigate the mechanics of curved fluid interfaces: Angle of Contact and Excess Pressure. We examine why curved liquid surfaces experience a pressure discontinuity, rigorously derive the Young-Laplace equation for liquid droplets, air cavities, and soap bubbles, analyze the geometry of coalescing bubble interfaces (Plateau’s laws), and deconstruct the wetting criteria governed by the Young-Dupré relation.
1. Detailed Concept Card
1.1 Curvature of Liquid Surfaces & Pressure Discontinuity
A planar liquid surface is in mechanical equilibrium under atmospheric pressure because surface tension forces along the surface resolve into zero net normal force. However, when an interface is curved:
- Concave Surface: The resultant of surface tension forces acting along the curved boundary points inward toward the center of curvature (toward the concave side).
- To maintain mechanical equilibrium across the boundary, the fluid pressure on the concave side must exceed the pressure on the convex side: $$\mathbf{P_{\text{concave}} > P_{\text{convex}}}$$
- The difference between the pressure on the concave side and the convex side is termed the Excess Pressure ($\Delta P = P_{\text{concave}} – P_{\text{convex}}$).
1.2 Derivation of Excess Pressure (The Young-Laplace Equation)
We derive the excess pressure using the Principle of Virtual Work. Consider expanding a curved surface by an infinitesimal virtual displacement $dR$ in equilibrium: the mechanical work done by the excess pressure force ($dW = \Delta P \cdot dV$) must equal the increase in surface interfacial energy ($dU_s = T \cdot dA$).
A. Spherical Liquid Drop (Single Interface in Gas)
A liquid drop of radius $R$ has a single liquid-gas interface. Increasing its radius by $dR$:
$$dV = 4\pi R^2 dR, \quad dA = 8\pi R dR$$
Equating virtual work to the increase in surface energy:
$$\Delta P \cdot dV = T \cdot dA \implies \Delta P (4\pi R^2 dR) = T (8\pi R dR)$$
$$\mathbf{\Delta P_{\text{drop}} = \frac{2T}{R}}$$
The interior pressure is higher than the outside ambient pressure: $P_{\text{in}} = P_0 + \frac{2T}{R}$.
B. Air Cavity / Bubble Inside a Liquid (Single Interface)
For an air bubble of radius $R$ formed within a liquid of surface tension $T$, there is only one interface separating the trapped air from the surrounding liquid. Hence:
$$\mathbf{\Delta P_{\text{cavity}} = \frac{2T}{R}}$$
If the air bubble is located at a depth $h$ below the free surface of a liquid of density $\rho$, the absolute pressure inside the bubble is:
$$P_{\text{in}} = P_0 + \rho g h + \frac{2T}{R}$$
C. Soap Bubble in Air (Two Free Interfaces)
A soap bubble in air consists of a thin spherical liquid film having two free interfaces (inner surface and outer surface). Both surfaces contribute equally to surface energy:
$$dA_{\text{total}} = 2 \times (8\pi R dR) = 16\pi R dR$$
$$\Delta P (4\pi R^2 dR) = T (16\pi R dR) \implies \mathbf{\Delta P_{\text{bubble}} = \frac{4T}{R}}$$
Thus, for identical radii and surface tension, a soap bubble in air sustains twice the excess pressure of a liquid droplet.
D. Cylindrical Liquid Jet (Single Curvature)
For a cylindrical jet of radius $R$ and length $L$, the interface curves in one dimension ($R_1 = R, R_2 = \infty$). The excess pressure is:
$$\mathbf{\Delta P_{\text{cylinder}} = \frac{T}{R}}$$
E. General Young-Laplace Equation for Arbitrary Interfaces
For any arbitrary smooth curved interface with principal orthogonal radii of curvature $R_1$ and $R_2$:
$$\mathbf{\Delta P = T \left(\frac{1}{R_1} + \frac{1}{R_2}\right)}$$
Special Case (Saddle / Minimal Surface): If $R_1 = -R_2$ (equal and opposite curvatures, such as in a catenoid soap film), $\Delta P = 0$. The pressure on both sides of the film is identical despite the surface being curved.
1.3 Common Interface of Two Contacting Soap Bubbles
Consider two spherical soap bubbles $1$ and $2$ of radii $r_1$ and $r_2$ with $r_1 < r_2$, blown from the same soap solution (surface tension $T$) in ambient air ($P_0$).
- Internal pressure of smaller bubble 1: $P_1 = P_0 + \frac{4T}{r_1}$.
- Internal pressure of larger bubble 2: $P_2 = P_0 + \frac{4T}{r_2}$.
- Since $r_1 < r_2$, it follows that $P_1 > P_2$. The smaller bubble has a higher internal pressure!
- When they coalesce into contact, the common partition film experiences a pressure difference: $$\Delta P_{\text{common}} = P_1 – P_2 = 4T\left(\frac{1}{r_1} – \frac{1}{r_2}\right)$$
- Let $r$ be the radius of curvature of this spherical common interface. Because it is a soap film with two surfaces: $$\frac{4T}{r} = 4T\left(\frac{1}{r_1} – \frac{1}{r_2}\right) \implies \mathbf{\frac{1}{r} = \frac{1}{r_1} – \frac{1}{r_2} \implies r = \frac{r_1 r_2}{r_2 – r_1}}$$
- Curvature Direction: Because $P_1 > P_2$, the common interface bulges convex into the larger bubble (its center of curvature lies inside the smaller bubble 1).
- Plateau’s Law of Tri-Junction Angles: The three soap films (bubble 1 outer film, bubble 2 outer film, and the common partition) meet along a circular rim at mutually symmetric angles of $120^\circ$ to achieve vector force equilibrium.
1.4 Angle of Contact ($\theta_c$) & Wetting Thermodynamics
The Angle of Contact $\theta$ is defined as the angle enclosed between the tangent to the liquid surface at the point of contact and the solid surface inside the liquid.
At the three-phase contact line (solid, liquid, vapor), equilibrium of interfacial surface tensions is governed by the Young-Dupré Equation:
$$\gamma_{SV} = \gamma_{SL} + \gamma_{LV} \cos\theta \implies \mathbf{\cos\theta = \frac{\gamma_{SV} – \gamma_{SL}}{\gamma_{LV}}}$$
where $\gamma_{SV}$ is solid-vapor interfacial energy, $\gamma_{SL}$ is solid-liquid interfacial energy, and $\gamma_{LV} = T$ is liquid-vapor surface tension.
Classification of Contact Angles & Wetting Regimes:
- 1. Acute Angle of Contact ($\theta < 90^\circ$):
- Condition: $\gamma_{SV} > \gamma_{SL}$ (adhesive force dominates: $F_a > F_c/\sqrt{2}$).
- Behavior: The liquid wets the solid surface.
- Meniscus: Concave in a glass capillary tube.
- Examples: Clean water on glass ($\theta \approx 0^\circ$ to $8^\circ$), kerosene on glass ($\theta \approx 26^\circ$).
- 2. Right Angle of Contact ($\theta = 90^\circ$):
- Condition: $\gamma_{SV} = \gamma_{SL} \implies F_a = F_c/\sqrt{2}$.
- Behavior: Neutral wetting.
- Meniscus: Completely flat.
- Example: Pure water on clean silver.
- 3. Obtuse Angle of Contact ($\theta > 90^\circ$):
- Condition: $\gamma_{SV} < \gamma_{SL}$ (cohesive force dominates: $F_a < F_c/\sqrt{2}$).
- Behavior: The liquid does not wet the solid surface; it beads up into droplets.
- Meniscus: Convex in a glass tube.
- Example: Mercury on glass ($\theta \approx 135^\circ$ to $140^\circ$), water on lotus leaf / wax.
- 4. Complete Spreading ($\theta = 0^\circ$):
- Condition: $\gamma_{SV} \ge \gamma_{SL} + \gamma_{LV}$. The liquid spreads spontaneously as a thin macroscopic film.
1.5 Critical Conceptual Insights & JEE Traps
- Trap 1: Air Bubble inside Liquid vs Soap Bubble in Air: An air bubble immersed in water has ONLY ONE interface. Its excess pressure is $\frac{2T}{R}$, NOT $\frac{4T}{R}$. Only a soap bubble in air (with both interior and exterior liquid-air surfaces) carries the factor of 4.
- Trap 2: Connecting Two Bubbles via a Valve: If a smaller soap bubble ($r_1$) and a larger soap bubble ($r_2$) are connected by a hollow pipe with an open stopcock, air flows from the smaller bubble into the larger bubble because $P_1 = P_0 + \frac{4T}{r_1} > P_2 = P_0 + \frac{4T}{r_2}$. The small bubble shrinks and collapses, while the large bubble grows!
- Trap 3: Contact Angle Independence of Tube Tilt: The contact angle $\theta$ is an intrinsic thermodynamic material property of the solid-liquid-gas triplet. Tilting the tube or altering tube radius does NOT alter the angle of contact $\theta$.
- Trap 4: Effect of Detergents: Detergents dramatically reduce $\gamma_{LV}$ and $\gamma_{SL}$, increasing $\cos\theta$ toward $1$ ($\theta \to 0^\circ$). This enhances wetting, allowing water to penetrate fabric micro-pores.
2. Four Graded Solved Examples
Example 1: Absolute & Excess Pressure of an Underwater Air Cavity (JEE Main Level)
Problem Statement: An air bubble of radius $r = 0.10\text{ mm}$ is formed at a depth $h = 10.0\text{ cm}$ below the free surface of water in an open container. The surface tension of water is $T = 0.075\text{ N/m}$, water density is $\rho = 1000\text{ kg/m}^3$, atmospheric pressure is $P_0 = 1.013 \times 10^5\text{ N/m}^2$, and $g = 9.8\text{ m/s}^2$. Calculate:
- (a) The excess pressure across the boundary of the air bubble.
- (b) The hydrostatic pressure in the water adjacent to the bubble.
- (c) The total absolute pressure inside the air bubble.
Step-by-Step Solution:
Step 1: Calculate the excess pressure $\Delta P$:
Because the air bubble is submerged in water, it possesses a single curved spherical liquid-air interface:
$$\Delta P = \frac{2T}{r}$$
Convert radius to SI units ($r = 0.10\text{ mm} = 1.0 \times 10^{-4}\text{ m}$):
$$\Delta P = \frac{2 \times 0.075\text{ N/m}}{1.0 \times 10^{-4}\text{ m}} = \frac{0.150}{10^{-4}} = 1500\text{ N/m}^2 = 1.50\text{ kPa}$$
Step 2: Calculate the external hydrostatic pressure $P_h$ at depth $h$:
$$P_h = P_0 + \rho g h = 1.013 \times 10^5 + (1000) \times (9.8) \times (0.10)$$
$$P_h = 101300 + 980 = 102280\text{ N/m}^2 = 102.28\text{ kPa}$$
Step 3: Calculate the total absolute pressure inside the bubble:
The interior of the bubble lies on the concave side of the interface, so its pressure exceeds $P_h$ by $\Delta P$:
$$P_{\text{in}} = P_h + \Delta P = 102280 + 1500 = 103780\text{ N/m}^2 = 103.78\text{ kPa}$$
Key Takeaway: Submerged gas bubbles carry an excess pressure of $\frac{2T}{r}$ (single interface), which adds directly onto the local depth hydrostatic pressure $P_0 + \rho gh$.
Example 2: Geometry of Common Partition Between Two Contacting Soap Bubbles
Problem Statement: Two soap bubbles of radii $r_1 = 3.0\text{ cm}$ and $r_2 = 6.0\text{ cm}$ blown from the same soap solution adhere together to form a double bubble sharing a common spherical partition film. Taking ambient atmospheric pressure as $P_0$ and surface tension as $T$:
- (a) Determine the radius of curvature $r$ of the common interface.
- (b) State which bubble the common interface bulges into, with justification.
- (c) Find the angle between the tangent planes of the three soap films at their mutual junction line.
Step-by-Step Solution:
Step 1: Write internal pressure equations for both bubbles:
$$P_1 = P_0 + \frac{4T}{r_1}, \quad P_2 = P_0 + \frac{4T}{r_2}$$
Since $r_1 = 3.0\text{ cm} < r_2 = 6.0\text{ cm}$, $P_1 > P_2$. The pressure difference across the common interface is:
$$\Delta P_{\text{common}} = P_1 – P_2 = 4T \left(\frac{1}{r_1} – \frac{1}{r_2}\right)$$
Step 2: Relate pressure difference to radius of curvature $r$:
Because the common interface is a soap film separating air of bubble 1 from air of bubble 2, it has two liquid-air surfaces:
$$\frac{4T}{r} = 4T \left(\frac{1}{r_1} – \frac{1}{r_2}\right) \implies \frac{1}{r} = \frac{1}{r_1} – \frac{1}{r_2} = \frac{r_2 – r_1}{r_1 r_2}$$
$$r = \frac{r_1 r_2}{r_2 – r_1} = \frac{3.0 \times 6.0}{6.0 – 3.0} = \frac{18.0}{3.0} = 6.0\text{ cm}$$
Step 3: Determine the bulge direction:
Because $P_1 > P_2$, the pressure is greater in the smaller bubble 1. Consequently, the common film must be concave toward bubble 1 and bulges convex into the larger bubble 2.
Step 4: Junction angle by Plateau’s Law:
At the circular triple junction line, each of the three soap films exerts an identical surface tension force per unit length ($2T$). For the line element to be in vector force equilibrium:
$$\vec{F}_1 + \vec{F}_2 + \vec{F}_3 = 0$$
Three coplanar equal vectors summing to zero must be mutually separated by equal angles of:
$$\theta = \frac{360^\circ}{3} = 120^\circ$$
Key Takeaway: The common interface radius is given by the reciprocal harmonic difference $r = \frac{r_1 r_2}{r_2 – r_1} = 6.0\text{ cm}$, always bulging toward the larger bubble, with triple-film junction angles strictly equal to $120^\circ$.
Example 3: Young-Dupré Relation & Contact Angle Thermodynamics (JEE Advanced Level)
Problem Statement: A droplet of liquid is deposited on a polished horizontal solid substrate in air. The interfacial surface energies are measured as $\gamma_{SV} = 0.080\text{ N/m}$ (solid-vapor), $\gamma_{SL} = 0.030\text{ N/m}$ (solid-liquid), and the surface tension of the liquid is $\gamma_{LV} = 0.065\text{ N/m}$.
- (a) Calculate the equilibrium contact angle $\theta$. Does the liquid wet the surface?
- (b) If a surfactant is dissolved in the liquid such that $\gamma_{LV}$ drops to $0.045\text{ N/m}$ and $\gamma_{SL}$ drops to $0.020\text{ N/m}$, determine the new contact angle.
- (c) Derive the general thermodynamic condition under which complete spontaneous spreading occurs ($\theta \to 0^\circ$).
Step-by-Step Solution:
Step 1: Calculate contact angle from the Young-Dupré equation:
$$\gamma_{SV} = \gamma_{SL} + \gamma_{LV}\cos\theta \implies \cos\theta = \frac{\gamma_{SV} – \gamma_{SL}}{\gamma_{LV}}$$
Substitute initial values:
$$\cos\theta = \frac{0.080 – 0.030}{0.065} = \frac{0.050}{0.065} = \frac{10}{13} \approx 0.7692$$
$$\theta = \arccos\left(\frac{10}{13}\right) \approx 39.7^\circ$$
Because $\theta < 90^\circ$, the liquid partially wets the solid surface.
Step 2: Effect of adding surfactant:
With the surfactant present, $\gamma_{SV} = 0.080\text{ N/m}$, $\gamma_{SL} = 0.020\text{ N/m}$, and $\gamma_{LV} = 0.045\text{ N/m}$:
$$\frac{\gamma_{SV} – \gamma_{SL}}{\gamma_{LV}} = \frac{0.080 – 0.020}{0.045} = \frac{0.060}{0.045} = \frac{4}{3} \approx 1.333$$
Since $\frac{\gamma_{SV} – \gamma_{SL}}{\gamma_{LV}} > 1$, a finite contact angle cannot satisfy horizontal equilibrium! The solid-vapor pulling force exceeds the combined resisting interfacial forces:
$$\gamma_{SV} > \gamma_{SL} + \gamma_{LV}$$
Consequently, the liquid droplet spontaneously spreads across the entire substrate into a continuous thin monolayer film: $\mathbf{\theta = 0^\circ}$ (Complete Wetting).
Step 3: Thermodynamic spreading condition:
The Spreading Coefficient $S$ is defined as:
$$S = \gamma_{SV} – (\gamma_{SL} + \gamma_{LV})$$
- If $S < 0$: Partial wetting occurs with a stable finite contact angle $\theta = \arccos\left(\frac{\gamma_{SV} – \gamma_{SL}}{\gamma_{LV}}\right)$.
- If $S \ge 0$: Spontaneous complete spreading occurs with $\theta = 0^\circ$.
Key Takeaway: Surfactants lower both $\gamma_{LV}$ and $\gamma_{SL}$, driving the spreading coefficient $S \ge 0$, which causes the contact angle to collapse to zero for maximum detergent cleaning action.
Example 4: Pressure-Volume Virtual Work Equivalence in Soap Bubble Growth
Problem Statement: Prove from first principles that the mechanical work done by internal gas pressure $\int \Delta P \, dV$ in growing a soap bubble from initial radius $R_1$ to final radius $R_2$ against surface tension $T$ is mathematically identical to the total increase in surface energy $\Delta U_s = T \cdot \Delta A_{\text{total}}$.
Step-by-Step Solution:
Step 1: Express excess pressure and differential volume:
For a soap bubble of instantaneous radius $R$ in ambient air, the excess pressure across the double-sided film is:
$$\Delta P(R) = \frac{4T}{R}$$
The differential change in internal volume as radius expands by $dR$ is:
$$dV = d\left(\frac{4}{3}\pi R^3\right) = 4\pi R^2 dR$$
Step 2: Set up and evaluate the work integral:
$$W = \int_{R_1}^{R_2} \Delta P(R) \, dV = \int_{R_1}^{R_2} \left(\frac{4T}{R}\right)(4\pi R^2 dR) = 16\pi T \int_{R_1}^{R_2} R \, dR$$
Evaluating the integral:
$$W = 16\pi T \left[ \frac{R^2}{2} \right]_{R_1}^{R_2} = 8\pi T (R_2^2 – R_1^2)$$
Step 3: Compare with total surface energy increase:
A soap bubble has two free spherical interfaces (inner and outer). The total initial and final surface areas are:
$$A_{\text{total}, 1} = 2 \times (4\pi R_1^2) = 8\pi R_1^2$$
$$A_{\text{total}, 2} = 2 \times (4\pi R_2^2) = 8\pi R_2^2$$
$$\Delta A_{\text{total}} = 8\pi (R_2^2 – R_1^2)$$
The total increase in surface potential energy is:
$$\Delta U_s = T \cdot \Delta A_{\text{total}} = 8\pi T (R_2^2 – R_1^2)$$
Thus, $W = \int \Delta P \, dV = \Delta U_s$ holds identically. Q.E.D.
Key Takeaway: The factor of $4$ in the Young-Laplace bubble formula $\Delta P = \frac{4T}{R}$ perfectly integrates over volume to yield the factor of $8\pi$ in double-sided surface energy expansion ($8\pi R^2 T$).
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. The excess pressure inside a spherical mercury droplet of radius $R = 2.0\text{ mm}$ (surface tension $T = 0.465\text{ N/m}$) exposed to air is:
- (A) $232.5\text{ Pa}$
- (B) $465.0\text{ Pa}$
- (C) $930.0\text{ Pa}$
- (D) $116.25\text{ Pa}$
Q2. Two spherical soap bubbles $A$ and $B$ have radii in the ratio $r_A : r_B = 2 : 3$. The ratio of their internal excess pressures $\Delta P_A : \Delta P_B$ is:
- (A) $4 : 9$
- (B) $2 : 3$
- (C) $3 : 2$
- (D) $9 : 4$
Q3. An air bubble of radius $r$ is situated at a depth $h$ below the free surface of a lake. If atmospheric pressure is $P_0$, water density is $\rho$, and surface tension is $T$, the total absolute pressure of air inside the bubble is:
- (A) $P_0 + \rho g h + \frac{4T}{r}$
- (B) $P_0 + \rho g h + \frac{2T}{r}$
- (C) $P_0 + \frac{2T}{r}$
- (D) $\rho g h + \frac{2T}{r}$
Q4. For a liquid-solid interface where adhesive forces greatly exceed cohesive forces ($F_a > F_c/\sqrt{2}$), the angle of contact and meniscus curvature are respectively:
- (A) Obtuse; Convex
- (B) Acute; Concave
- (C) Acute; Convex
- (D) Zero; Flat
Q5. Two soap bubbles of radii $4.0\text{ cm}$ and $6.0\text{ cm}$ come into contact to form a shared partition film. The radius of curvature of this common interface is:
- (A) $2.4\text{ cm}$
- (B) $5.0\text{ cm}$
- (C) $10.0\text{ cm}$
- (D) $12.0\text{ cm}$
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. Which of the following statements regarding excess pressure and surface curvature are correct?
- (A) The pressure on the concave side of a curved fluid boundary is always greater than on the convex side.
- (B) The excess pressure inside a cylindrical liquid jet of radius $R$ is $\Delta P = \frac{T}{R}$.
- (C) For a saddle-shaped minimal surface where principal radii of curvature satisfy $R_1 = -R_2$, the net excess pressure is zero.
- (D) The excess pressure inside an air bubble in water of radius $R$ is equal to the excess pressure inside a soap bubble in air of the same radius.
Q7. Regarding the angle of contact $\theta$ and interfacial wetting, which of the following statements are correct?
- (A) The Young-Dupré equation gives $\cos\theta = \frac{\gamma_{SV} – \gamma_{SL}}{\gamma_{LV}}$.
- (B) If $\gamma_{SV} < \gamma_{SL}$, the contact angle is obtuse and the liquid does not wet the solid.
- (C) Adding a surfactant reduces the surface tension $\gamma_{LV}$, increasing $\cos\theta$ and facilitating spreading.
- (D) The angle of contact depends directly on the angle of inclination of the solid surface with respect to the vertical.
Q8. Two spherical soap bubbles of radii $r_1$ and $r_2$ ($r_1 < r_2$) meet and share a common boundary film. Which of the following statements are correct?
- (A) The common boundary film is spherical and bulges into the larger bubble.
- (B) The radius of curvature of the common film is $r = \frac{r_1 r_2}{r_2 – r_1}$.
- (C) If $r_1 = r_2$, the common boundary film is planar ($r \to \infty$).
- (D) The angle between each adjacent pair of films at the junction line is $90^\circ$.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. Two soap bubbles of radii $r_1 = 2.0\text{ cm}$ and $r_2 = 3.0\text{ cm}$ coalesce into contact. The radius of curvature of their common internal interface is $r$ (in $\text{cm}$). Find the value of $r$.
Q10. A narrow cylindrical water jet issuing from a nozzle has a radius $R = 1.0\text{ mm}$ and surface tension $T = 0.075\text{ N/m}$. Calculate the excess pressure (in $\text{Pa}$) inside the cylindrical jet.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (B) | Excess Pressure in Liquid Drop |
| Q2 | (C) | Inverse Radius Ratio of Excess Pressure |
| Q3 | (B) | Absolute Pressure in Underwater Cavity |
| Q4 | (B) | Adhesion Dominance & Meniscus Shape |
| Q5 | (D) | Common Soap Bubble Interface Radius |
| Q6 | (A), (B), (C) | Young-Laplace Curvature Principles |
| Q7 | (A), (B), (C) | Young-Dupré Wetting Criteria |
| Q8 | (A), (B), (C) | Coalescent Bubble Film Geometry |
| Q9 | 6 | Common Interface Radius Calculation |
| Q10 | 75 | Excess Pressure in Cylindrical Jet |
Step-by-Step Solutions
Solution 1:
A mercury drop in air is a single liquid-gas interface. The excess pressure is:
$$\Delta P = \frac{2T}{R} = \frac{2 \times 0.465\text{ N/m}}{2.0 \times 10^{-3}\text{ m}} = \frac{0.930}{2.0 \times 10^{-3}} = 465.0\text{ Pa}$$
Correct Option: (B).
Solution 2:
For soap bubbles in air: $\Delta P = \frac{4T}{r}$.
$$\frac{\Delta P_A}{\Delta P_B} = \frac{r_B}{r_A} = \frac{3}{2} \implies 3 : 2$$
Correct Option: (C).
Solution 3:
At depth $h$, the external hydrostatic pressure in the liquid surrounding the bubble is $P_h = P_0 + \rho g h$. The bubble has a single interface, so the pressure inside exceeds $P_h$ by $\frac{2T}{r}$:
$$P_{\text{in}} = P_0 + \rho g h + \frac{2T}{r}$$
Correct Option: (B).
Solution 4:
When adhesive forces exceed cohesive forces ($F_a > F_c/\sqrt{2}$), the resultant force vector on boundary molecules points into the solid wall. Equilibrium demands that the liquid surface aligns perpendicular to this resultant, tilting upward at the boundary. Hence, the angle of contact is acute ($\theta < 90^\circ$) and the meniscus is concave.
Correct Option: (B).
Solution 5:
The common interface radius satisfies:
$$r = \frac{r_1 r_2}{r_2 – r_1} = \frac{4.0 \times 6.0}{6.0 – 4.0} = \frac{24.0}{2.0} = 12.0\text{ cm}$$
Correct Option: (D).
Solution 6:
- (A) is Correct: The resultant of surface tension forces along the perimeter pulls toward the center of curvature (concave side), requiring higher pressure on the concave side for balance.
- (B) is Correct: For a cylinder, curvature along the length is zero ($R_2 = \infty$), yielding $\Delta P = T(1/R + 0) = \frac{T}{R}$.
- (C) is Correct: By Young-Laplace, $\Delta P = T(1/R_1 + 1/R_2)$. If $R_1 = -R_2$, $\Delta P = 0$.
- (D) is Incorrect: Air bubble in water has one interface ($\Delta P = 2T/R$), whereas soap bubble in air has two interfaces ($\Delta P = 4T/R$).
Correct Options: (A), (B), (C).
Solution 7:
- (A) is Correct: This is the standard Young-Dupré equation derived from three-phase interfacial line tension balance.
- (B) is Correct: If $\gamma_{SV} < \gamma_{SL}$, $\cos\theta < 0$, which implies $90^\circ < \theta \le 180^\circ$ (non-wetting).
- (C) is Correct: Surfactants lower $\gamma_{LV}$, making $\cos\theta$ more positive and driving $\theta$ toward $0^\circ$.
- (D) is Incorrect: Contact angle is an intrinsic material property independent of tube orientation or geometric tilt.
Correct Options: (A), (B), (C).
Solution 8:
- (A) is Correct: Since $P_1 > P_2$, the net pressure force pushes the partition film into bubble 2 (convex toward bubble 2, concave toward bubble 1).
- (B) is Correct: Derived from $\frac{4T}{r} = \frac{4T}{r_1} – \frac{4T}{r_2} \implies r = \frac{r_1 r_2}{r_2 – r_1}$.
- (C) is Correct: When $r_1 = r_2$, $P_1 = P_2 \implies \Delta P = 0$, so the common partition is completely flat ($r = \infty$).
- (D) is Incorrect: Plateau’s law requires symmetric three-force balance: the angle between any pair of films is $120^\circ$, not $90^\circ$.
Correct Options: (A), (B), (C).
Solution 9:
Using the common radius formula:
$$r = \frac{r_1 r_2}{r_2 – r_1} = \frac{2.0 \times 3.0}{3.0 – 2.0} = \frac{6.0}{1.0} = 6.0\text{ cm}$$
Final Answer: 6
Solution 10:
For a cylindrical jet of radius $R = 1.0\text{ mm} = 1.0 \times 10^{-3}\text{ m}$:
$$\Delta P = \frac{T}{R} = \frac{0.075\text{ N/m}}{1.0 \times 10^{-3}\text{ m}} = 75\text{ Pa}$$
Final Answer: 75
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