Welcome to Lecture 88 of our comprehensive JEE Physics Master Series. In this lecture, we examine one of the most elegant interfacial manifestations of fluid mechanics: Capillary Action and Applications of Surface Tension. We deconstruct Jurin’s law through both force balance and Laplace pressure deficit frameworks, analyze the non-overflow theorem for tubes of insufficient length, establish the $50\%$ thermal dissipation theorem in capillary ascent, and investigate capillary rise between parallel plates and wedge geometries.
1. Detailed Concept Card
1.1 Mechanism of Capillarity & Derivation of Jurin’s Law
When a narrow bore glass tube (capillary tube, typically internal radius $r < 1\text{ mm}$) open at both ends is dipped vertically into a wetting liquid such as water, the liquid spontaneously ascends inside the tube above the outside free surface. Conversely, for a non-wetting liquid such as mercury, the liquid level inside the tube is depressed below the surrounding level. This phenomenon is termed Capillarity.
Derivation Method 1: Upward Surface Tension Force Balance
Let a capillary tube of internal radius $r$ dip vertically into a liquid of density $\rho$ and surface tension $T$. Let the contact angle between the liquid and the inner glass wall be $\theta$.
- The boundary of contact between the liquid meniscus and the tube wall is a circle of circumference $2\pi r$.
- Along this entire circular contact perimeter, surface tension pulls obliquely downward on the glass with force $T$ per unit length at an angle $\theta$ to the wall.
- By Newton’s third law, the glass wall pulls upward on the liquid perimeter with an equal and opposite force directed along the tangent to the meniscus. The net vertical upward force supporting the liquid column is: $$F_v = (2\pi r) T \cos\theta$$
- This upward surface tension force supports the total weight $W$ of the elevated liquid column of height $h$: $$W = m g = \rho V g = \rho (\pi r^2 h) g$$
- Equating upward pulling force to downward gravitational weight ($F_v = W$): $$(2\pi r) T \cos\theta = \rho \pi r^2 h g$$
Solving for the capillary rise height $h$ yields Jurin’s Law:
$$\mathbf{h = \frac{2T\cos\theta}{\rho g r}}$$
Derivation Method 2: Laplace Pressure Deficit Across the Meniscus
Consider the curved spherical meniscus of radius of curvature $R$. From elementary geometry of the circular tube bore:
$$R = \frac{r}{\cos\theta}$$
Because the meniscus is concave on the upper (air) side, the pressure in the liquid immediately below the meniscus ($P_A$) is lower than atmospheric pressure $P_0$ by the Young-Laplace excess pressure:
$$P_A = P_0 – \frac{2T}{R} = P_0 – \frac{2T\cos\theta}{r}$$
At the flat free liquid surface outside the tube, pressure is atmospheric ($P_0$). At the same horizontal level inside the tube at depth $h$ below the meniscus:
$$P = P_A + \rho g h = P_0 – \frac{2T\cos\theta}{r} + \rho g h$$
Hydrostatic equilibrium across communicating vessels requires that points at the same horizontal level in a continuous fluid have identical pressure ($P = P_0$):
$$P_0 – \frac{2T\cos\theta}{r} + \rho g h = P_0 \implies \rho g h = \frac{2T\cos\theta}{r} \implies h = \frac{2T\cos\theta}{\rho g r}$$
Meniscus Volume Correction
For complete wetting ($\theta \approx 0^\circ$), the meniscus is a hemisphere of radius $r$. The liquid trapped in the meniscus above height $h$ occupies a volume:
$$V_{\text{meniscus}} = V_{\text{cylinder}} – V_{\text{hemisphere}} = \pi r^2(r) – \frac{2}{3}\pi r^3 = \frac{1}{3}\pi r^3$$
The effective height of the liquid column is $h_{\text{eff}} = h + \frac{r}{3}$. The exact corrected Jurin formula is:
$$h + \frac{r}{3} = \frac{2T}{\rho g r} \implies \mathbf{h = \frac{2T}{\rho g r} – \frac{r}{3}}$$
1.2 Capillary Tube of Insufficient Length ($l < h$)
A classic conceptual trap in JEE is: “What happens if a capillary tube is cut to a length $l$ that is less than the theoretical capillary rise $h$ ($l < h$)?”
- The Non-Overflow Theorem: The liquid never overflows, fountains, or drips from the top rim. If it were to overflow continuously, it would constitute a perpetual motion machine operating without external energy input!
- Meniscus Flattening Adjustment: As the liquid column reaches the upper rim of height $l$, the contact line pins at the sharp lip. The meniscus flattens, increasing its radius of curvature from $R$ to a larger radius $R’$ ($R’ > R$).
- The apparent contact angle increases from $\theta$ to $\theta’$ ($\theta’ > \theta$) such that the hydrostatic pressure of the shorter column balances the reduced Laplace pressure deficit: $$\rho g l = \frac{2T}{R’} = \frac{2T\cos\theta’}{r}$$
- Equating with the full-height condition $\rho g h = \frac{2T}{R}$: $$\mathbf{h \cdot R = l \cdot R’ = \text{constant} = \frac{2T}{\rho g}}$$
- The new radius of curvature and new contact angle are: $$\mathbf{R’ = \left(\frac{h}{l}\right) R, \quad \cos\theta’ = \left(\frac{l}{h}\right) \cos\theta}$$
- Limiting Case: If $l \to 0$, $R’ \to \infty$, and the meniscus becomes completely flat ($\theta’ = 90^\circ$).
1.3 Inclined Capillary Tubes
If a capillary tube is tilted at an angle $\alpha$ with respect to the vertical:
- The vertical height $h$ of the liquid meniscus above the free surface remains completely unchanged: $$h = \frac{2T\cos\theta}{\rho g r}$$
- The length of the liquid column $L$ measured along the slanted tube bore increases: $$\mathbf{L = \frac{h}{\cos\alpha} = \frac{2T\cos\theta}{\rho g r \cos\alpha}}$$
1.4 Work, Potential Energy & Thermal Dissipation during Ascent
Consider the energetic balance of the liquid ascending from rest to equilibrium height $h$ in a vertical capillary tube ($\theta = 0^\circ$):
- 1. Work Done by Surface Tension Force ($W$): The constant upward perimeter force $F = 2\pi r T$ acts over displacement $h$: $$W = F \cdot h = (2\pi r T) h = (\rho \pi r^2 h g) h = \rho \pi r^2 h^2 g$$
- 2. Gravitational Potential Energy Gained by Liquid ($U_g$): The center of mass of the uniform liquid column of mass $m = \rho \pi r^2 h$ rises by $h/2$: $$U_g = m g \left(\frac{h}{2}\right) = (\rho \pi r^2 h) g \left(\frac{h}{2}\right) = \frac{1}{2}\rho \pi r^2 h^2 g = \mathbf{\frac{1}{2} W}$$
- 3. Energy Dissipated as Heat ($Q_{\text{heat}}$): By the First Law of Thermodynamics, the difference between the mechanical work performed by surface tension and the stored potential energy is dissipated as heat via viscous fluid friction during the ascent: $$Q_{\text{heat}} = W – U_g = \mathbf{\frac{1}{2} W = \frac{1}{2}\rho \pi r^2 h^2 g}$$
Fundamental Law: In any spontaneous capillary rise, exactly $50\%$ of the work done by surface tension is dissipated as heat, while the remaining $50\%$ is stored as gravitational potential energy.
1.5 Capillary Rise Between Parallel Plates & Wedge Films
A. Two Vertical Parallel Glass Plates Separated by Gap $d$
Consider two vertical glass plates of width $L$ placed parallel to each other with a narrow gap $d$ ($d \ll L$) dipping into water:
- The upward surface tension force acts along the width $L$ on both plates: $$F_v = 2 \times (T L \cos\theta)$$
- The weight of the raised liquid sheet of thickness $d$, width $L$, and height $h$ is: $$W = m g = \rho (L d h) g$$
- Equating force and weight: $$2 T L \cos\theta = \rho L d h g \implies \mathbf{h = \frac{2T\cos\theta}{\rho g d}}$$
B. Wedge-Shaped Glass Plates (Hyperbolic Meniscus Profile)
If two vertical glass plates touch at one vertical edge and are inclined at a very small angle $\beta$, the separation between them at horizontal distance $x$ from the contact apex is $d(x) = \beta x$. The capillary rise height at coordinate $x$ is:
$$h(x) = \frac{2T\cos\theta}{\rho g \beta x} \implies \mathbf{h(x) \cdot x = \text{constant}}$$
The profile of the elevated water curve between the plates forms a rectangular hyperbola.
1.6 Critical Conceptual Insights & JEE Traps
- Trap 1: Weightlessness in Orbit: In an orbiting satellite or free fall ($g_{\text{eff}} = 0$), the denominator of Jurin’s formula vanishes ($h \to \infty$). The liquid rises immediately to the very top of the tube, no matter how long the tube is, and the meniscus flattens to a plane ($R’ \to \infty$).
- Trap 2: Mass vs Radius Proportionality: While height varies inversely with radius ($h \propto 1/r$), the mass of liquid elevated varies directly with radius: $$m = \rho \pi r^2 h = \rho \pi r^2 \left(\frac{2T\cos\theta}{\rho g r}\right) = \frac{2\pi T\cos\theta}{g} r \implies \mathbf{m \propto r}$$
- Trap 3: Capillary Depression: For mercury in glass, $\theta \approx 135^\circ > 90^\circ$, so $\cos\theta < 0$. The liquid level inside the tube is depressed ($h < 0$) and the meniscus is convex.
- Trap 4: Pressure in the Capillary Column: The pressure inside the liquid column at any height $y$ above the free surface is $P(y) = P_0 – \rho g y$. It is strictly sub-atmospheric throughout the entire elevated column!
2. Four Graded Solved Examples
Example 1: Capillary Rise & Meniscus Volume Correction (JEE Main Level)
Problem Statement: A clean vertical glass capillary tube of internal bore radius $r = 0.25\text{ mm}$ is dipped into water of surface tension $T = 0.075\text{ N/m}$ and density $\rho = 1000\text{ kg/m}^3$. Taking $g = 9.8\text{ m/s}^2$ and contact angle $\theta = 0^\circ$:
- (a) Calculate the capillary rise height $h$ using the standard Jurin formula.
- (b) Calculate the corrected height $h_{\text{corr}}$ accounting for the volume of water in the hemispherical meniscus.
- (c) Find the percentage error in ignoring the meniscus correction.
Step-by-Step Solution:
Step 1: Calculate standard Jurin’s rise $h$:
Convert radius to meters: $r = 0.25\text{ mm} = 2.5 \times 10^{-4}\text{ m}$.
$$h = \\frac{2 T \cos\theta}{\rho g r} = \frac{2 \times 0.075 \times 1.0}{1000 \times 9.8 \times (2.5 \times 10^{-4})} = \frac{0.150}{2.45} \approx 0.06122\text{ m} = 6.122\text{ cm}$$
Step 2: Apply the hemispherical meniscus correction:
The meniscus correction subtracts $r/3$ from the standard Jurin height:
$$\frac{r}{3} = \frac{0.25\text{ mm}}{3} \approx 0.0833\text{ mm} = 0.00833\text{ cm}$$
$$h_{\text{corr}} = h – \frac{r}{3} = 6.1224 – 0.0083 = 6.1141\text{ cm}$$
Step 3: Calculate percentage error:
$$\text{Error (\%)} = \frac{r/3}{h_{\text{corr}}} \times 100 = \frac{0.00833}{6.1141} \times 100 \approx 0.136\%$$
Key Takeaway: For narrow tubes with $r \ll h$, the meniscus correction is well under $0.2\%$, completely justifying the standard approximation $h = \frac{2T}{\rho gr}$.
Example 2: Capillary Tube of Insufficient Length & Meniscus Adjustment
Problem Statement: A glass capillary tube of radius $r$ dipped in water shows a normal capillary rise of $h = 8.0\text{ cm}$ with zero contact angle ($\theta = 0^\circ$). The tube is now pulled out and cut so that its length above the outside water surface is only $l = 5.0\text{ cm}$. When re-immersed vertically:
- (a) Will water overflow from the upper tip? Explain physically.
- (b) Determine the new radius of curvature $R’$ of the water meniscus in terms of $r$.
- (c) Calculate the new apparent contact angle $\theta’$.
Step-by-Step Solution:
Step 1: Physical explanation of non-overflow:
Water will not overflow. As water reaches the upper rim of the tube, the boundary line of contact is arrested at the sharp edge. Rather than spilling over, the liquid surface adjusts its curvature, becoming less concave to balance the smaller hydrostatic head.
Step 2: Calculate new radius of curvature $R’$:
In the original full-height tube, the meniscus is hemispherical with radius of curvature $R = r / \cos(0^\circ) = r$. The hydrostatic pressure balance is:
$$\rho g h = \frac{2T}{R}$$
In the truncated tube of length $l = 5.0\text{ cm}$:
$$\rho g l = \frac{2T}{R’}$$
Dividing the two equations:
$$\frac{l}{h} = \frac{R}{R’} \implies R’ = \left(\frac{h}{l}\right) R = \left(\frac{8.0\text{ cm}}{5.0\text{ cm}}\right) r = 1.60\, r$$
Step 3: Calculate the new apparent contact angle $\theta’$:
From the geometric relation $R’ = \frac{r}{\cos\theta’}$:
$$\cos\theta’ = \frac{r}{R’} = \frac{l}{h} = \frac{5.0}{8.0} = 0.625$$
$$\theta’ = \arccos(0.625) \approx 51.32^\circ$$
Key Takeaway: The product $l \cdot R’ = h \cdot R$ is an invariant constant. A shorter tube simply forces the meniscus to flatten, increasing $R’$ to $1.6r$ and opening the apparent contact angle to $51.3^\circ$.
Example 3: Thermodynamic Energy Balance & Heat Dissipation (JEE Advanced Level)
Problem Statement: A capillary tube of radius $r = 0.50\text{ mm}$ is held vertically and touched to the surface of water ($T = 0.070\text{ N/m}$, $\rho = 1000\text{ kg/m}^3$). Take $g = 10\text{ m/s}^2$ and $\theta = 0^\circ$. As water ascends to its equilibrium height:
- (a) Calculate the equilibrium height $h$.
- (b) Calculate the total work $W$ performed by the surface tension force.
- (c) Calculate the gravitational potential energy $U_g$ gained by the elevated liquid column.
- (d) Determine the quantity of heat $Q_{\text{heat}}$ dissipated during the spontaneous ascent.
Step-by-Step Solution:
Step 1: Calculate the equilibrium height $h$:
$$r = 0.50\text{ mm} = 5.0 \times 10^{-4}\text{ m}$$
$$h = \frac{2 T \cos\theta}{\rho g r} = \frac{2 \times 0.070 \times 1.0}{1000 \times 10 \times (5.0 \times 10^{-4})} = \frac{0.140}{5.0} = 0.028\text{ m} = 2.80\text{ cm}$$
Step 2: Calculate work done by surface tension $W$:
$$F = 2\pi r T = 2\pi \times (5.0 \times 10^{-4}) \times 0.070 = 7.0\pi \times 10^{-5}\text{ N}$$
$$W = F \cdot h = (7.0\pi \times 10^{-5}\text{ N}) \times (0.028\text{ m}) = 1.96\pi \times 10^{-6}\text{ J} \approx 6.158 \times 10^{-6}\text{ J} = 6.16\text{ }\mu\text{J}$$
Step 3: Calculate gravitational potential energy $U_g$:
Mass of the column: $m = \rho \pi r^2 h = 1000 \times \pi \times (5.0 \times 10^{-4})^2 \times 0.028 = 7.0\pi \times 10^{-6}\text{ kg}$.
$$U_g = m g \left(\frac{h}{2}\right) = (7.0\pi \times 10^{-6}) \times 10 \times \left(\frac{0.028}{2}\right) = 0.98\pi \times 10^{-6}\text{ J} \approx 3.079 \times 10^{-6}\text{ J} = 3.08\text{ }\mu\text{J}$$
Notice that $U_g = \frac{1}{2} W$ exactly!
Step 4: Calculate heat dissipated $Q_{\text{heat}}$:
$$Q_{\text{heat}} = W – U_g = 6.158\text{ }\mu\text{J} – 3.079\text{ }\mu\text{J} = 3.079\text{ }\mu\text{J} \approx 3.08\text{ }\mu\text{J}$$
Key Takeaway: Exactly $50\%$ of the work done by surface tension is dissipated as thermal energy through fluid viscous drag, proving that capillary ascent is inherently dissipative.
Example 4: Capillary Rise & Attractive Force Between Parallel Glass Plates
Problem Statement: Two flat, clean vertical glass plates of width $L = 10.0\text{ cm}$ are positioned parallel to each other with a narrow separation $d = 0.50\text{ mm}$. The lower edges of the plates dip into a trough of water ($T = 0.075\text{ N/m}$, $\rho = 1000\text{ kg/m}^3$, $\theta = 0^\circ$, $g = 10\text{ m/s}^2$).
- (a) Calculate the height $h$ to which water rises between the plates.
- (b) Calculate the net inward attractive force pulling the two glass plates together.
Step-by-Step Solution:
Step 1: Calculate the height of capillary rise $h$:
The gap is $d = 0.50\text{ mm} = 5.0 \times 10^{-4}\text{ m}$. Using the parallel plate formula:
$$h = \frac{2 T \cos\theta}{\rho g d} = \frac{2 \times 0.075 \times 1.0}{1000 \times 10 \times (5.0 \times 10^{-4})} = \frac{0.150}{5.0} = 0.030\text{ m} = 3.0\text{ cm}$$
Step 2: Calculate the net inward force pressing the plates together:
At any elevation $y$ above the free surface ($0 \le y \le h$), the pressure in the water film between the plates is sub-atmospheric: $P(y) = P_0 – \rho g y$.
The atmospheric pressure $P_0$ pressing inward from the dry outer faces exceeds the water pressure on the inner faces by $\Delta P(y) = \rho g y$. The average pressure deficit across the elevated liquid area is:
$$\overline{\Delta P} = \frac{1}{2} \rho g h$$
The area of the wetted plate surface is $A_{\text{wetted}} = L \cdot h$. Thus, the total inward force on each plate is:
$$F = \overline{\Delta P} \cdot A_{\text{wetted}} = \left(\frac{1}{2}\rho g h\right)(L h) = \frac{1}{2}\rho g L h^2$$
Substitute $h = \frac{2T}{\rho g d}$:
$$F = \frac{1}{2}\rho g L \left(\frac{2T}{\rho g d}\right)^2 = \frac{2 T^2 L}{\rho g d^2}$$
Substitute numerical values:
$$F = \frac{1}{2} \times (1000) \times (10) \times (0.10\text{ m}) \times (0.030\text{ m})^2 = 500 \times 0.10 \times 0.0009 = 0.450\text{ N}$$
Key Takeaway: Two wet glass plates adhere together with strong force ($0.45\text{ N}$) due to the negative gauge pressure ($\overline{\Delta P} = 150\text{ Pa}$) sustained across the elevated capillary water film.
3. Practice Worksheet (10 Questions)
Questions 1 to 5: JEE Main Level (Single Correct Option)
Q1. Water rises to a height $h$ in a glass capillary tube of radius $r$. If the tube is replaced by another tube of radius $r/2$, the new height of water column will be:
- (A) $h/2$
- (B) $h$
- (C) $2h$
- (D) $4h$
Q2. A capillary tube of radius $r$ dipped vertically in water shows a rise of $h = 4.0\text{ cm}$. If the tube is inclined at an angle of $60^\circ$ with the vertical, the length of the water column along the tube will be:
- (A) $2.0\text{ cm}$
- (B) $4.0\text{ cm}$
- (C) $8.0\text{ cm}$
- (D) $4\sqrt{3}\text{ cm}$
Q3. A glass capillary tube of radius $r = 1.0\text{ mm}$ is dipped vertically into mercury ($T = 0.49\text{ N/m}$, $\rho = 13600\text{ kg/m}^3$, $\theta = 135^\circ$). Taking $g = 9.8\text{ m/s}^2$ and $\cos(135^\circ) = -\frac{1}{\sqrt{2}}$, the capillary depression of mercury inside the tube is approximately:
- (A) $-2.6\text{ mm}$
- (B) $-5.2\text{ mm}$
- (C) $-10.4\text{ mm}$
- (D) $-1.3\text{ mm}$
Q4. The mass $m$ of liquid supported inside a vertical capillary tube of internal radius $r$ varies with $r$ as:
- (A) $m \propto \frac{1}{r}$
- (B) $m \propto r^0$ (independent of $r$)
- (C) $m \propto r$
- (D) $m \propto r^2$
Q5. A clean glass capillary tube of length $20\text{ cm}$ is taken inside an artificial Earth satellite in a stable orbit. When its lower end is dipped into a water beaker:
- (A) Water will not rise into the tube at all.
- (B) Water will rise to its normal terrestrial height ($h \approx 6\text{ cm}$).
- (C) Water will rise to the very top ($20\text{ cm}$) and overflow continuously.
- (D) Water will rise to the very top ($20\text{ cm}$) and form a flat meniscus without overflowing.
Questions 6 to 8: JEE Advanced Level (One or More than One Correct)
Q6. A capillary tube of radius $r$ has a normal capillary rise $h$. The tube is cut to length $l < h$ and immersed vertically into the liquid. Let $R$ be the original radius of curvature of the meniscus and $R’$ be the new radius of curvature. Which of the following statements are correct?
- (A) The liquid does not overflow from the top edge.
- (B) The new radius of curvature is $R’ = \left(\frac{h}{l}\right) R$.
- (C) The apparent contact angle $\theta’$ satisfies $\cos\theta’ = \left(\frac{l}{h}\right) \cos\theta$.
- (D) The product of liquid column height and meniscus radius of curvature remains invariant: $l \cdot R’ = h \cdot R$.
Q7. During the spontaneous ascent of a liquid column of height $h$ in a vertical capillary tube of radius $r$ (zero contact angle), which of the following statements are correct?
- (A) The total work done by surface tension is $W = \rho \pi r^2 h^2 g$.
- (B) The gravitational potential energy stored in the liquid column is $U = \frac{1}{2} \rho \pi r^2 h^2 g$.
- (C) Exactly $50\%$ of the work performed by surface tension is converted into heat.
- (D) If the tube is frictionless (zero viscosity), the liquid column would execute simple harmonic motion about the equilibrium height $h$.
Q8. Two identical clean glass plates of width $L$ are placed vertically in water with a narrow separation. Which of the following statements are correct?
- (A) For parallel plates separated by distance $d$, the height of capillary rise is $h = \frac{2T\cos\theta}{\rho g d}$.
- (B) For parallel plates, the net attractive force pulling the plates together is $F = \frac{2T^2 L}{\rho g d^2}$.
- (C) If the plates form a wedge of small angle $\beta$, the meniscus profile $h(x)$ along horizontal distance $x$ from the apex is a parabola.
- (D) For wedge plates, the curve formed by the liquid meniscus is a rectangular hyperbola: $h(x) \cdot x = \text{constant}$.
Questions 9 to 10: Numerical / Integer Answer Type
Q9. Water rises to an equilibrium height $h = 10.0\text{ cm}$ in a vertical capillary tube with a hemispherical meniscus of radius of curvature $R$. If the tube is cut at a height $l = 2.5\text{ cm}$ above the water surface, the new radius of curvature of the meniscus at the top becomes $R’ = K \cdot R$. Find the integer value of $K$.
Q10. Water rises to a height $h_1 = 40.0\text{ mm}$ in a vertical glass capillary tube of internal radius $r_1 = 0.50\text{ mm}$. In another vertical capillary tube of internal radius $r_2 = 0.20\text{ mm}$ dipped in the same water vessel, water rises to a height $h_2$ (in $\text{mm}$). Calculate the value of $h_2$.
4. Detailed Solutions & Answer Key
Quick Answer Key
| Question | Correct Answer | Topic / Category |
|---|---|---|
| Q1 | (C) | Jurin’s Law Radius Scaling |
| Q2 | (C) | Inclined Tube Length |
| Q3 | (B) | Capillary Depression of Mercury |
| Q4 | (C) | Elevated Mass Proportionality |
| Q5 | (D) | Zero-Gravity Meniscus Behavior |
| Q6 | (A), (B), (C), (D) | Insufficient Length Non-Overflow Dynamics |
| Q7 | (A), (B), (C), (D) | Energy & Dissipation in Capillary Rise |
| Q8 | (A), (B), (D) | Capillary Action Between Plates |
| Q9 | 4 | Meniscus Curvature Scaling Ratio |
| Q10 | 100 | Capillary Rise Inverse Radius Calculation |
Step-by-Step Solutions
Solution 1:
By Jurin’s Law, $h = \frac{2T\cos\theta}{\rho g r} \implies h \cdot r = \text{constant}$.
$$h_2 = h_1 \left(\frac{r_1}{r_2}\right) = h \left(\frac{r}{r/2}\right) = 2h$$
Correct Option: (C).
Solution 2:
The vertical height $h = 4.0\text{ cm}$ remains constant. For an inclination angle $\alpha = 60^\circ$ with the vertical:
$$L = \frac{h}{\cos\alpha} = \frac{4.0\text{ cm}}{\cos(60^\circ)} = \frac{4.0}{0.5} = 8.0\text{ cm}$$
Correct Option: (C).
Solution 3:
$$h = \frac{2 T \cos\theta}{\rho g r} = \frac{2 \times 0.49 \times (-0.7071)}{13600 \times 9.8 \times (1.0 \times 10^{-3})} = \frac{-0.69296}{133.28} \approx -0.00520\text{ m} = -5.2\text{ mm}$$
The negative sign signifies capillary depression. Correct Option: (B).
Solution 4:
The total upward surface tension force supporting the column is $F = 2\pi r T \cos\theta$. In equilibrium, $m g = F = 2\pi r T \cos\theta$, which directly implies:
$$m = \left(\frac{2\pi T \cos\theta}{g}\right) r \implies m \propto r$$
Correct Option: (C).
Solution 5:
In a satellite in free fall, $g_{\text{eff}} = 0$. Jurin’s formula gives $h \to \infty$. Therefore, water ascends to fill the entire tube ($20\text{ cm}$). When it reaches the top, the meniscus flattens completely ($R’ \to \infty$) so that excess Laplace pressure becomes zero, arresting the flow without spilling over.
Correct Option: (D).
Solution 6:
- (A) is Correct: Spilling over would violate conservation of energy.
- (B) is Correct: $\rho g l = \frac{2T}{R’} \implies R’ = \frac{h}{l} R$.
- (C) is Correct: $\cos\theta’ = \frac{r}{R’} = \frac{l}{h}\cos\theta$.
- (D) is Correct: $\rho g (l R’) = 2T = \rho g (h R) \implies l R’ = h R$.
Correct Options: (A), (B), (C), (D).
Solution 7:
- (A) is Correct: $W = (2\pi r T) h = (\rho \pi r^2 h g) h = \rho \pi r^2 h^2 g$.
- (B) is Correct: Center of mass rises to $h/2$, so $U = m g (h/2) = \frac{1}{2}\rho \pi r^2 h^2 g$.
- (C) is Correct: $Q_{\text{heat}} = W – U = \frac{1}{2} W$ ($50\%$ dissipation).
- (D) is Correct: In the absence of viscous dissipation, the excess energy would drive undamped harmonic oscillations about $h$.
Correct Options: (A), (B), (C), (D).
Solution 8:
- (A) is Correct: $2 T L \cos\theta = \rho L d h g \implies h = \frac{2T\cos\theta}{\rho g d}$.
- (B) is Correct: $F = \overline{\Delta P} \cdot A = \frac{1}{2}\rho g h \cdot (L h) = \frac{1}{2}\rho g L h^2 = \frac{2 T^2 L}{\rho g d^2}$.
- (C) is Incorrect: $h(x) \propto 1/x$ is a hyperbola, not a parabola.
- (D) is Correct: $d(x) = \beta x \implies h(x) = \frac{2T\cos\theta}{\rho g \beta x} \implies h(x) \cdot x = \text{constant}$ (rectangular hyperbola).
Correct Options: (A), (B), (D).
Solution 9:
Using the invariant product $l \cdot R’ = h \cdot R$:
$$R’ = \left(\frac{h}{l}\right) R = \left(\frac{10.0\text{ cm}}{2.5\text{ cm}}\right) R = 4 R \implies K = 4$$
Final Answer: 4
Solution 10:
By Jurin’s law, $h_1 r_1 = h_2 r_2$:
$$h_2 = h_1 \left(\frac{r_1}{r_2}\right) = (40.0\text{ mm}) \times \left(\frac{0.50\text{ mm}}{0.20\text{ mm}}\right) = 40.0 \times 2.5 = 100.0\text{ mm}$$
Final Answer: 100
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