Welcome to Lecture 86 of our comprehensive JEE Physics Master Series. In this lecture, we explore the molecular and mechanical foundations of interfacial phenomena: Surface Tension and Surface Energy. We deconstruct Laplace’s molecular theory of cohesion, derive the work-energy relations for stretching liquid interfaces and thin films, evaluate the thermodynamics of droplet breaking and coalescence, and establish the governing laws for wire-frame equilibria.
1. Detailed Concept Card
1.1 Molecular Basis of Surface Tension: Laplace’s Theory
Intermolecular forces in liquids are classified into two categories:
- Cohesive Forces: Attractive forces between molecules of the same substance (e.g., water-water molecular attraction).
- Adhesive Forces: Attractive forces between molecules of different substances (e.g., water-glass attraction).
The maximum distance up to which a molecule can exert significant attraction on neighboring molecules is called the molecular range ($c \approx 10^{-9}\text{ m} = 1\text{ nm}$). An imaginary sphere drawn around a molecule with radius equal to this molecular range is called its sphere of molecular influence.
Consider molecules located at different depths within a liquid:
- 1. Deep Interior Molecule (Bulk): Completely surrounded by neighboring liquid molecules in all directions. The net cohesive force acting on it is symmetrically balanced and identically zero ($\vec{F}_{\text{net}} = 0$).
- 2. Molecule in the Surface Film: The surface film is the top layer of liquid whose thickness equals the molecular range $c$. For a molecule lying at the free surface, its lower hemisphere lies entirely in liquid, whereas its upper hemisphere is exposed to air (containing far fewer molecules). Consequently, it experiences an unbalanced net downward cohesive pull directed perpendicularly into the bulk liquid.
Because every surface molecule is drawn inward, the free surface of a liquid is under constant tension and tends to minimize its surface area, behaving like a stretched elastic membrane. This macroscopic phenomenon is known as Surface Tension.
1.2 Mechanical Definition of Surface Tension ($T$ or $S$)
Imagine an imaginary line of length $L$ drawn tangentially across the liquid surface. The fluid on either side of this line exerts a pulling force on the fluid on the opposite side, directed perpendicular to the line and tangential to the liquid surface.
The Surface Tension $T$ is defined as the tangential force per unit length acting perpendicular to this imaginary line:
$$T = \frac{F}{L}$$
- SI Unit: $\text{Newton per meter } (\text{N/m})$ or $\text{Joule per square meter } (\text{J/m}^2)$.
- CGS Unit: $\text{dyne/cm} \quad (1\text{ N/m} = 10^3\text{ dyne/cm})$.
- Dimensional Formula: $[T] = \frac{[\text{M}^1 \text{L}^1 \text{T}^{-2}]}{[\text{L}^1]} = [\mathbf{M^1 L^0 T^{-2}}]$.
1.3 Surface Energy ($U_s$) & Work Done in Area Expansion
To bring a molecule from the interior bulk to the surface film, work must be done against the inward attractive cohesive forces. This work is stored as excess potential energy in the surface layer, termed Surface Energy ($U_s$).
Consider a rectangular wire frame with a movable frictionless wire slider of length $L$. When dipped into a soap solution, a thin soap film is formed. Because a liquid film has two free interfaces (top and bottom) exposed to air, the inward pulling force exerted by surface tension on the slider is:
$$F = 2 \cdot T \cdot L$$
To pull the slider outward quasi-statically by a displacement $\Delta x$, the work done by an external force is:
$$W = F \cdot \Delta x = (2 T L) \Delta x = T (2 L \Delta x) = T \cdot \Delta A_{\text{total}}$$
where $\Delta A_{\text{total}} = 2 L \Delta x$ is the total increase in surface area across both free surfaces. Hence, the surface tension is numerically and dimensionally equal to the surface energy per unit area:
$$\mathbf{T = \frac{W}{\Delta A_{\text{total}}} = \frac{\Delta U_s}{\Delta A_{\text{total}}}}$$
Crucial Surface Count Distinction:
- Single Interface (Liquid Drops, Liquid-Gas Meniscus): Only one free surface. Work done in expanding surface area by $\Delta A$ is: $$W = T \cdot \Delta A$$
- Double Interface (Soap Bubbles, Liquid Films in air): Two distinct free surfaces. Work done in expanding each planar face by $\Delta A$ is: $$W = T \cdot (2 \Delta A) = 2 T \Delta A$$
1.4 Splitting & Coalescence of Liquid Droplets
A. Atomization / Splitting of a Large Drop into $n$ Droplets
Consider a single spherical liquid drop of radius $R$ shattered into $n$ identical tiny droplets of radius $r$. Assuming incompressibility, volume conservation gives:
$$\frac{4}{3}\pi R^3 = n \left(\frac{4}{3}\pi r^3\right) \implies R^3 = n r^3 \implies r = \frac{R}{n^{1/3}}$$
The initial and final surface areas are:
$$A_i = 4\pi R^2, \quad A_f = n (4\pi r^2) = n \cdot 4\pi \left(\frac{R}{n^{1/3}}\right)^2 = 4\pi R^2 n^{1/3}$$
The net change in surface area is strictly positive ($\Delta A > 0$):
$$\Delta A = A_f – A_i = 4\pi R^2 \left(n^{1/3} – 1\right) = 4\pi R^3 \left(\frac{1}{r} – \frac{1}{R}\right)$$
The work done on the system to break the drop is:
$$W = T \cdot \Delta A = 4\pi T R^2 \left(n^{1/3} – 1\right) = 3 V T \left(\frac{1}{r} – \frac{1}{R}\right)$$
If no external heat is supplied (adiabatic condition), the required surface energy is extracted from the internal thermal energy of the liquid, resulting in a fall in temperature (cooling effect):
$$\Delta \theta = -\frac{W}{m \cdot s} = -\frac{3 V T \left(\frac{1}{r} – \frac{1}{R}\right)}{\rho V s} = -\frac{3T}{\rho s}\left(\frac{1}{r} – \frac{1}{R}\right)$$
B. Coalescence of $n$ Droplets into a Single Large Drop
Conversely, when $n$ small droplets coalesce to form one large drop, the total surface area decreases ($\Delta A < 0$). Surface energy is released in the form of heat, leading to a temperature rise:
$$\Delta Q = 4\pi T R^2 (n^{1/3} – 1) \implies \Delta \theta = +\frac{3T}{\rho s}\left(\frac{1}{r} – \frac{1}{R}\right)$$
1.5 Factors Influencing Surface Tension
- 1. Temperature Dependence: Surface tension decreases with increasing temperature because elevated kinetic energy weakens intermolecular cohesive attractions. At the critical temperature ($T_c$), the meniscus between liquid and vapor disappears, and surface tension becomes identically zero ($T = 0$). Ferguson’s Empirical Formula: $T = T_0 \left(1 – \frac{\theta}{\theta_c}\right)^n$ (where $n \approx 1.2$ for most liquids).
- 2. Soluble Impurities:
- Highly Soluble Substances (e.g., NaCl / Sugar in water): Increase surface tension due to strong ion-dipole attractive forces.
- Sparingly Soluble Substances & Surfactants (e.g., Soap, Detergents, Phenol): Drastically decrease surface tension because surfactant molecules congregate at the interface, disrupting water-water hydrogen bonding.
- 3. Surface Contamination: Dust, oil, or grease films on the surface significantly reduce the surface tension of water.
- 4. Electrification: Charging a liquid surface reduces its effective surface tension due to mutual electrostatic repulsion among surface charges.
1.6 Critical Conceptual Insights & JEE Traps
- Trap 1: Forgetting the factor of 2 in soap films: Always inspect whether the problem involves a liquid drop (1 surface, $W = T \cdot 4\pi R^2$) or a soap bubble / soap film (2 surfaces, $W = 2 \cdot T \cdot 4\pi R^2 = 8\pi R^2 T$).
- Trap 2: Direction of Surface Tension: Surface tension acts tangentially along the liquid interface. It is NEVER directed normal (perpendicular) to the surface; normal pressure differences arise solely from interface curvature (Laplace excess pressure).
- Trap 3: Work in pulling a submerged ring vs disc:
- A solid thin disc lifted from water has ONE contact perimeter: $F = T \cdot (2\pi R)$.
- A thin ring lifted from water has TWO contact perimeters (inner and outer): $F = T \cdot (2\pi R_1 + 2\pi R_2) \approx 4\pi R T$.
- Trap 4: Kinetic Energy in Coalescence: In rapid droplet coalescence without thermal dissipation, released surface energy is converted entirely into kinetic energy: $\frac{1}{2} M v^2 = \Delta U_s \implies v = \sqrt{\frac{6T}{\rho}\left(\frac{1}{r} – \frac{1}{R}\right)}$.
2. Four Graded Solved Examples
Example 1: Work Done in Expanding a Soap Bubble (JEE Main Level)
Problem Statement: A soap bubble is blown in air using a soap solution having a surface tension $T = 0.030\text{ N/m}$. Calculate the mechanical work required to expand the radius of the soap bubble from $R_1 = 3.0\text{ cm}$ to $R_2 = 5.0\text{ cm}$. Take $\pi \approx 3.1416$.
Step-by-Step Solution:
Step 1: Identify the number of free interfaces:
A soap bubble consists of a thin spherical liquid film enclosing air within, while surrounded by ambient air on the outside. Therefore, it possesses two free interfaces (an inner surface and an outer surface).
Step 2: Calculate the change in surface area for each interface:
$$\Delta A_{\text{single}} = 4\pi R_2^2 – 4\pi R_1^2 = 4\pi (R_2^2 – R_1^2)$$
Converting radii to SI units ($R_1 = 0.03\text{ m}, R_2 = 0.05\text{ m}$):
$$R_2^2 – R_1^2 = (0.05)^2 – (0.03)^2 = 0.0025 – 0.0009 = 0.0016\text{ m}^2 = 1.6 \times 10^{-3}\text{ m}^2$$
$$\Delta A_{\text{single}} = 4\pi \times 1.6 \times 10^{-3} = 6.4\pi \times 10^{-3}\text{ m}^2$$
Step 3: Total change in surface area across both interfaces:
$$\Delta A_{\text{total}} = 2 \times \Delta A_{\text{single}} = 8\pi (R_2^2 – R_1^2) = 12.8\pi \times 10^{-3}\text{ m}^2$$
Step 4: Compute the work done $W$:
$$W = T \cdot \Delta A_{\text{total}} = 0.030 \times (12.8\pi \times 10^{-3})$$
$$W = 0.384\pi \times 10^{-3}\text{ J} = 3.84\pi \times 10^{-4}\text{ J} \approx 1.206 \times 10^{-3}\text{ J} = 1.21\text{ mJ}$$
Key Takeaway: For any hollow liquid bubble in gas, the surface area factor is $8\pi(R_2^2 – R_1^2)$, whereas for a solid liquid drop it would be $4\pi(R_2^2 – R_1^2)$.
Example 2: Temperature Drop During Droplet Atomization (Thermodynamic Linkage)
Problem Statement: A spherical water drop of radius $R = 1.0\text{ mm}$ is sprayed adiabatically into $n = 10^6$ identical tiny spherical droplets. Given that the surface tension of water is $T = 0.072\text{ N/m}$, density $\rho = 1000\text{ kg/m}^3$, and specific heat capacity $s = 4200\text{ J/(kg}\cdot\text{K)}$, determine:
- (a) The radius $r$ of each newly formed droplet.
- (b) The mechanical work $W$ required to disperse the drop.
- (c) The resulting drop in temperature $\Delta \theta$ of the water mist.
Step-by-Step Solution:
Step 1: Determine the daughter droplet radius $r$:
By conservation of total liquid volume:
$$\frac{4}{3}\pi R^3 = n \left(\frac{4}{3}\pi r^3\right) \implies r = \frac{R}{n^{1/3}} = \frac{1.0 \times 10^{-3}\text{ m}}{(10^6)^{1/3}} = \frac{1.0 \times 10^{-3}}{100} = 1.0 \times 10^{-5}\text{ m} = 10\text{ }\mu\text{m}$$
Step 2: Calculate work done $W$:
Since water drops have only a single free liquid-air interface:
$$W = T \cdot \Delta A = T [n(4\pi r^2) – 4\pi R^2] = 4\pi T R^2 [n^{1/3} – 1]$$
Substitute numerical values:
$$4\pi R^2 = 4\pi (1.0 \times 10^{-3})^2 = 4\pi \times 10^{-6}\text{ m}^2$$
$$W = (4\pi \times 10^{-6}) \times 0.072 \times (100 – 1) = (4\pi \times 10^{-6}) \times 0.072 \times 99$$
$$W = 28.512\pi \times 10^{-6}\text{ J} \approx 8.957 \times 10^{-5}\text{ J}$$
Step 3: Calculate the adiabatic temperature drop:
The mass of the initial water drop is:
$$m = \rho V = \rho \left(\frac{4}{3}\pi R^3\right) = 1000 \times \frac{4}{3}\pi (10^{-3})^3 = \frac{4}{3}\pi \times 10^{-6}\text{ kg} \approx 4.189 \times 10^{-6}\text{ kg}$$
Under adiabatic conditions, this work is drawn from the internal energy of the droplets ($W = -m s \Delta \theta$):
$$\Delta \theta = -\frac{W}{m s} = -\frac{3 T}{\rho s}\left(\frac{1}{r} – \frac{1}{R}\right)$$
$$\frac{3T}{\rho s} = \frac{3 \times 0.072}{1000 \times 4200} = \frac{0.216}{4.2 \times 10^6} \approx 5.143 \times 10^{-8}\text{ K}\cdot\text{m}$$
$$\frac{1}{r} – \frac{1}{R} = \frac{1}{10^{-5}} – \frac{1}{10^{-3}} = 10^5 – 10^3 = 99000\text{ m}^{-1}$$
$$\Delta \theta = -(5.143 \times 10^{-8}) \times 99000 \approx -5.09 \times 10^{-3}\text{ K} = -5.09\text{ mK}$$
Key Takeaway: Atomization of liquids causes a distinct intrinsic drop in temperature ($5.09\text{ mK}$) due to the conversion of thermal internal energy into interfacial surface energy.
Example 3: Vertical Wire Frame Equilibrium & Sliding Film Dynamics (JEE Advanced Level)
Problem Statement: A U-shaped rigid wire frame is held vertically inside a container. A lightweight horizontal wire slider of length $L = 10.0\text{ cm}$ can slide freely along the two vertical arms without friction. A thin soap film ($T = 0.030\text{ N/m}$) is formed within the frame, supporting the slider against gravity.
- (a) Taking $g = 10\text{ m/s}^2$, find the mass $m$ of the slider required for static equilibrium.
- (b) If an external vertical force pulls the slider downward quasi-statically by a displacement $d = 5.0\text{ cm}$, calculate the total work done by the external pulling force.
- (c) If the slider is released from rest at this displaced position, determine its initial upward acceleration.
Step-by-Step Solution:
Step 1: Establish force balance in static equilibrium:
The soap film has two free liquid-air interfaces. Each interface exerts an upward surface tension force $T L$ along the length of the slider. Thus, the net upward surface tension force is:
$$F_{\text{up}} = 2 T L = 2 \times 0.030\text{ N/m} \times 0.10\text{ m} = 0.0060\text{ N} = 6.0 \times 10^{-3}\text{ N}$$
For static equilibrium, downward weight must equal upward surface tension:
$$m g = F_{\text{up}} = 2 T L \implies m = \frac{2 T L}{g} = \frac{0.0060}{10} = 6.0 \times 10^{-4}\text{ kg} = 0.60\text{ g}$$
Step 2: Calculate external work done during downward displacement:
During quasi-static motion, the external applied force balances the net restoring force. Because gravity assists downward motion, the external force required is:
$$F_{\text{ext}} + m g = 2 T L \implies F_{\text{ext}} = 2 T L – m g$$
Since $m g = 2 T L$ throughout (surface tension force is independent of film length $x$), $F_{\text{ext}} = 0$. The gravitational potential energy lost by the slider precisely supplies the required increase in surface energy:
$$\Delta U_s = 2 T L d = 2 \times 0.030 \times 0.10 \times 0.05 = 3.0 \times 10^{-4}\text{ J}$$
$$\Delta U_g = -m g d = -(6.0 \times 10^{-4}) \times 10 \times 0.05 = -3.0 \times 10^{-4}\text{ J}$$
$$W_{\text{ext}} = \Delta U_s + \Delta U_g = 0\text{ J}$$
Note: If the frame were placed horizontally, gravity does not act along the direction of motion, and the external work would be $W = 2 T L d = 3.0 \times 10^{-4}\text{ J} = 0.30\text{ mJ}$.
Step 3: Initial upward acceleration upon release:
At any position along the uniform arms, $F_{\text{net}} = 2 T L – m g = 0$. Hence, the slider remains in neutral equilibrium at any position, and its acceleration upon release is:
$$a = \frac{F_{\text{net}}}{m} = 0\text{ m/s}^2$$
Key Takeaway: Unlike a Hookean spring where restoring force increases with elongation ($F = -kx$), the surface tension force $F = 2TL$ is completely independent of the extension distance $x$, creating a neutral equilibrium state in a uniform vertical frame.
Example 4: Isothermal Coalescence of Two Soap Bubbles in Vacuum (Olympiad / JEE Advanced)
Problem Statement: Two spherical soap bubbles of radii $r_1 = 3.0\text{ cm}$ and $r_2 = 4.0\text{ cm}$ coalesce under strictly isothermal conditions in a complete vacuum to form a single larger spherical bubble of radius $R$. Assuming the soap solution has surface tension $T$ and the enclosed gas behaves as an ideal gas, prove that $R = \sqrt{r_1^2 + r_2^2}$ and find the numerical value of $R$.
Step-by-Step Solution:
Step 1: Express internal pressure of a soap bubble in vacuum:
In vacuum, ambient pressure is $P_0 = 0$. Because a soap bubble has two surfaces, the excess pressure across the bubble wall is:
$$P_{\text{in}} – P_0 = \frac{4T}{r} \implies P_1 = \frac{4T}{r_1}, \quad P_2 = \frac{4T}{r_2}$$
Step 2: Conservation of gas molecules under isothermal conditions:
Let $n_1$ and $n_2$ be the moles of air trapped inside the two bubbles. By the Ideal Gas Law ($P V = n R_{\text{gas}} \Theta$):
$$n_1 = \frac{P_1 V_1}{R_{\text{gas}} \Theta}, \quad n_2 = \frac{P_2 V_2}{R_{\text{gas}} \Theta}$$
When they coalesce into a single bubble of radius $R$ at the same temperature $\Theta$, total moles are conserved ($n_{\text{final}} = n_1 + n_2$):
$$P R^3 = P_1 r_1^3 + P_2 r_2^3 \implies P V = P_1 V_1 + P_2 V_2$$
Step 3: Substitute pressure and volume expressions:
$$P = \frac{4T}{R}, \quad V = \frac{4}{3}\pi R^3 \implies P V = \left(\frac{4T}{R}\right)\left(\frac{4}{3}\pi R^3\right) = \frac{16}{3}\pi T R^2$$
Similarly, for the initial bubbles:
$$P_1 V_1 = \frac{16}{3}\pi T r_1^2, \quad P_2 V_2 = \frac{16}{3}\pi T r_2^2$$
Equating the total $P V$ product:
$$\frac{16}{3}\pi T R^2 = \frac{16}{3}\pi T r_1^2 + \frac{16}{3}\pi T r_2^2$$
Dividing through by the common factor $\frac{16}{3}\pi T$:
$$\mathbf{R^2 = r_1^2 + r_2^2 \implies R = \sqrt{r_1^2 + r_2^2}}$$
Step 4: Calculate numerical value:
$$R = \sqrt{(3.0)^2 + (4.0)^2} = \sqrt{9.0 + 16.0} = \sqrt{25.0} = 5.0\text{ cm}$$
Key Takeaway: In vacuum, isothermal bubble coalescence leads to the Pythagorean sum of squared radii ($R^2 = r_1^2 + r_2^2$), directly reflecting that the total internal energy of trapped gas scales with the surface area.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. The dimensional formula for surface tension $T$ is identical to the dimensional formula of which of the following physical quantities?
- (A) Pressure
- (B) Force constant (spring constant)
- (C) Modulus of elasticity
- (D) Viscosity
Q2. A thin circular metal ring of mean radius $R = 5.0\text{ cm}$ rests horizontally on the surface of water ($T = 0.075\text{ N/m}$). The minimum additional upward force required to pull the ring cleanly off the water surface (neglecting the weight of the ring) is approximately:
- (A) $0.0236\text{ N}$
- (B) $0.0471\text{ N}$
- (C) $0.0942\text{ N}$
- (D) $0.0118\text{ N}$
Q3. The work done in blowing a spherical soap bubble of radius $R$ is $W_1$. The additional work required to further expand its radius from $R$ to $2R$ under isothermal conditions is:
- (A) $W_1$
- (B) $2 W_1$
- (C) $3 W_1$
- (D) $7 W_1$
Q4. When $1000$ identical spherical water droplets of radius $r = 1.0\text{ mm}$ coalesce into a single large spherical drop, the ratio of the initial total surface energy to the final surface energy is:
- (A) $100 : 1$
- (B) $10 : 1$
- (C) $1 : 10$
- (D) $1000 : 1$
Q5. Which of the following statements correctly describes the effect of temperature and solutes on the surface tension of water?
- (A) Surface tension increases with increasing temperature and becomes infinite at the critical temperature.
- (B) Adding common salt ($\text{NaCl}$) slightly decreases the surface tension of water.
- (C) Adding soap or detergent dramatically decreases the surface tension of water.
- (D) Surface tension of a liquid is completely independent of the nature of the ambient gas in contact with it.
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. A spherical liquid drop of radius $R$ is broken into $n$ identical tiny droplets of radius $r$ under adiabatic conditions. Which of the following statements are correct?
- (A) The radius of each tiny droplet is $r = R n^{-1/3}$.
- (B) The net work done on the drop is $W = 4\pi T R^2 (n^{1/3} – 1)$.
- (C) The process is accompanied by an increase in temperature of the liquid.
- (D) If the released energy is conserved entirely as kinetic energy, the speed of the droplets is $v = \sqrt{\frac{6T}{\rho}\left(\frac{1}{r} – \frac{1}{R}\right)}$.
Q7. Which of the following statements regarding the molecular theory of surface tension are correct?
- (A) Molecules in the surface film possess higher potential energy than molecules in the bulk interior.
- (B) The thickness of the surface film is of the order of the molecular range ($\sim 10^{-9}\text{ m}$).
- (C) Surface tension acts along the outward normal to the free liquid surface.
- (D) At the critical temperature, the surface tension of a liquid vanishes because the distinction between liquid and vapor phases disappears.
Q8. A rectangular wire frame dipped in a soap solution has a soap film formed on it. A movable slider of length $L$ is held in position. Which of the following statements are correct?
- (A) The force exerted by the film on the slider is $F = 2 T L$.
- (B) If the slider is moved by a distance $x$ parallel to itself, the work done against surface tension is $W = 2 T L x$.
- (C) The force required to hold the slider depends linearly on the current area of the film.
- (D) If a cotton thread loop is placed on the soap film and punctured inside, it assumes a circular shape to minimize the area of the remaining film.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. A horizontal slider wire of length $L = 10.0\text{ cm}$ on a soap film ($T = 0.030\text{ N/m}$) is displaced slowly by a distance of $\Delta x = 2.0\text{ cm}$ increasing the film area. Calculate the work done by the external pulling force in microjoules ($\mu\text{J}$).
Q10. When $64$ identical spherical liquid droplets coalesce into a single large drop, the ratio of the total initial surface energy to the final surface energy of the coalesced drop is equal to an integer $K$. Find the value of $K$.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (B) | Dimensional Analysis of Surface Tension |
| Q2 | (B) | Upward Pull on Floating Ring |
| Q3 | (C) | Isothermal Bubble Expansion Work |
| Q4 | (B) | Droplet Coalescence Energy Ratio |
| Q5 | (C) | Effects of Solutes and Surfactants |
| Q6 | (A), (B), (D) | Atomization Thermodynamics & Velocity |
| Q7 | (A), (B), (D) | Molecular Theory & Film Thickness |
| Q8 | (A), (B), (D) | Wire Frame Film Mechanics |
| Q9 | 120 | Work Done on Soap Film Slider |
| Q10 | 4 | Energy Ratio in Droplet Coalescence |
Step-by-Step Solutions
Solution 1:
Surface tension is defined as force per unit length: $[T] = \frac{[\text{M}^1 \text{L}^1 \text{T}^{-2}]}{[\text{L}^1]} = [\text{M}^1 \text{L}^0 \text{T}^{-2}]$.
- (A) Pressure: $[P] = [\text{M}^1 \text{L}^{-1} \text{T}^{-2}]$.
- (B) Force constant (spring constant $k = F/x$): $[k] = \frac{[\text{M}^1 \text{L}^1 \text{T}^{-2}]}{[\text{L}^1]} = [\text{M}^1 \text{L}^0 \text{T}^{-2}]$. Identical to surface tension!
- (C) Modulus of elasticity: $[Y] = [\text{M}^1 \text{L}^{-1} \text{T}^{-2}]$.
- (D) Dynamic Viscosity: $[\eta] = [\text{M}^1 \text{L}^{-1} \text{T}^{-1}]$.
Correct Option: (B).
Solution 2:
When a thin circular ring of radius $R$ is pulled away from a liquid surface, liquid adheres to both the inner perimeter and the outer perimeter of the ring.
Total contact perimeter is:
$$L_{\text{eff}} = 2\pi R_{\text{inner}} + 2\pi R_{\text{outer}} \approx 2(2\pi R) = 4\pi R$$
The total upward surface tension force required is:
$$F = T \cdot (4\pi R) = 0.075\text{ N/m} \times 4\pi \times (0.05\text{ m}) = 0.015\pi\text{ N} \approx 0.04712\text{ N}$$
Correct Option: (B).
Solution 3:
For a soap bubble of radius $R$ blown from a flat film:
$$W_1 = 2 \times 4\pi R^2 T = 8\pi R^2 T$$
To expand it from radius $R$ to $2R$, the additional work required is:
$$W_2 = 8\pi T \left[(2R)^2 – R^2\right] = 8\pi T [4R^2 – R^2] = 8\pi T (3R^2) = 3(8\pi R^2 T) = 3 W_1$$
Correct Option: (C).
Solution 4:
By volume conservation: $R = n^{1/3} r = (1000)^{1/3} r = 10 r$.
Initial surface energy of $n$ droplets:
$$U_i = n (4\pi r^2 T)$$
Final surface energy of the single coalesced drop:
$$U_f = 4\pi R^2 T = 4\pi (10r)^2 T = 100(4\pi r^2 T)$$
The ratio is:
$$\frac{U_i}{U_f} = \frac{n(4\pi r^2 T)}{100(4\pi r^2 T)} = \frac{1000}{100} = 10 \implies 10 : 1$$
Correct Option: (B).
Solution 5:
- (A) is False: Surface tension decreases with temperature and vanishes at the critical temperature.
- (B) is False: Highly soluble salts like $\text{NaCl}$ increase the surface tension of water.
- (C) is True: Surfactants (detergents/soaps) concentrate at the surface and severely lower surface tension, enabling water to wet fabrics better.
- (D) is False: Interfacial surface tension depends on the nature of both contacting media.
Correct Option: (C).
Solution 6:
- (A) is Correct: Incompressible volume conservation dictates $\frac{4}{3}\pi R^3 = n \left(\frac{4}{3}\pi r^3\right) \implies r = R n^{-1/3}$.
- (B) is Correct: $\Delta A = n(4\pi r^2) – 4\pi R^2 = 4\pi R^2 (n^{1/3} – 1) \implies W = T \Delta A = 4\pi T R^2(n^{1/3} – 1)$.
- (C) is Incorrect: Because surface area increases, energy is absorbed from internal thermal energy, causing cooling ($\Delta \theta < 0$).
- (D) is Correct: If the energy difference is entirely transformed into kinetic energy: $\frac{1}{2} M v^2 = W \implies \frac{1}{2} (\rho \frac{4}{3}\pi R^3) v^2 = 4\pi T R^3 (\frac{1}{r} – \frac{1}{R}) \implies v = \sqrt{\frac{6T}{\rho}\left(\frac{1}{r} – \frac{1}{R}\right)}$.
Correct Options: (A), (B), (D).
Solution 7:
- (A) is Correct: Surface molecules have an unshared coordination sphere and work was done to bring them to the surface against cohesive forces.
- (B) is Correct: The surface film thickness equals the molecular range of intermolecular attraction ($\sim 10^{-9}\text{ m}$).
- (C) is Incorrect: Surface tension acts tangentially along the surface plane, not along the normal.
- (D) is Correct: At $T_c$, latent heat of vaporization becomes zero and the meniscus disappears, so $T = 0$.
Correct Options: (A), (B), (D).
Solution 8:
- (A) is Correct: The soap film has two free surfaces, each pulling with force $TL$, so $F = 2TL$.
- (B) is Correct: $W = F \cdot x = (2TL)x = 2TLx$.
- (C) is Incorrect: The restoring force $F = 2TL$ depends only on the width $L$ of the slider, not on the total area of the film.
- (D) is Correct: When punctured, the remaining film pulls outward in all directions to minimize its own surface area, which maximizes the interior hole area into a perfect circle.
Correct Options: (A), (B), (D).
Solution 9:
The soap film has two free interfaces. The pulling force is:
$$F = 2 T L = 2 \times 0.030\text{ N/m} \times 0.10\text{ m} = 0.0060\text{ N}$$
The displacement is $\Delta x = 2.0\text{ cm} = 0.020\text{ m}$.
$$W = F \Delta x = 0.0060\text{ N} \times 0.020\text{ m} = 1.20 \times 10^{-4}\text{ J}$$
Converting to microjoules ($1\text{ J} = 10^6\text{ }\mu\text{J}$):
$$W = 1.20 \times 10^{-4} \times 10^6\text{ }\mu\text{J} = 120\text{ }\mu\text{J}$$
Final Answer: 120
Solution 10:
For $n = 64$ droplets coalescing into a single drop of radius $R$:
$$R = n^{1/3} r = (64)^{1/3} r = 4 r$$
Initial total surface energy of $64$ droplets:
$$U_i = 64 \times (4\pi r^2 T) = 256\pi r^2 T$$
Final surface energy of the coalesced drop:
$$U_f = 4\pi R^2 T = 4\pi (4r)^2 T = 4\pi (16 r^2) T = 64\pi r^2 T$$
The ratio is:
$$K = \frac{U_i}{U_f} = \frac{256\pi r^2 T}{64\pi r^2 T} = 4$$
Final Answer: 4