Maximum Height of Water Above Glass Rim Due to Surface Tension

To find the maximum height $h$ of the water above the glass rim just before overflowing, we can analyze the problem using two standard approaches: Pressure Balance Method (based on the problem statement's text) and Force Balance Method (based on horizontal force equilibrium).

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Approach 1: Pressure Balance Method

1. Hydrostatic Pressure at the Rim: The height of the water region above the glass rim is $h$. The excess hydrostatic pressure at the bottom of this region (at the rim level) relative to atmospheric pressure is: $\Delta P_{\text{hydrostatic}} = \rho g h$

2. Excess Pressure Supported by Surface Tension: The edge of the water disc is modeled as a cylindrical surface with a semicircular vertical profile of diameter $h$. Thus, the radius of curvature $R$ of the surface is: $R = \frac{h}{2}$

   The maximum excess pressure that can be supported by surface tension for a cylindrical surface of radius $R$ is given by the Young-Laplace equation: $\Delta P_{\text{surface tension}} = \frac{T}{R} = \frac{T}{h/2} = \frac{2T}{h}$

3. Condition for Overflow: Just before the water starts to overflow, the hydrostatic pressure at the bottom equals the maximum pressure supported by surface tension: $\Delta P_{\text{hydrostatic}} = \Delta P_{\text{surface tension}}$ $\rho g h = \frac{2T}{h}$ $h^2 = \frac{2T}{\rho g}$ $h = \sqrt{\frac{2T}{\rho g}}$

4. Calculation: Given values:

   • Density of water, $\rho = 10^3 \text{ kg m}^{-3}$
   • Surface tension, $T = 0.07 \text{ N m}^{-1}$
   • Acceleration due to gravity, $g = 10 \text{ m s}^{-2}$

   Substituting these values into the expression: $h = \sqrt{\frac{2 \times 0.07}{10^3 \times 10}} = \sqrt{\frac{0.14}{10^4}} = \sqrt{14 \times 10^{-6}} \text{ m}$ $h = \sqrt{14} \times 10^{-3} \text{ m} \approx 3.74 \text{ mm}$

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Approach 2: Force Balance Method

1. Hydrostatic Force: Consider a segment of length $L$ along the rim. The outward horizontal force on this segment due to the gauge hydrostatic pressure is: $F_{\text{hydrostatic}} = \int_{0}^{h} (\rho g y) L \, dy = \frac{1}{2} \rho g h^2 L$

2. Surface Tension Force: The inward horizontal force holding the liquid surface intact is exerted by surface tension at both the top and bottom boundaries of the semicircular edge: $F_{\text{surface tension}} = 2 T L$

3. Equilibrium Condition: Equating the outward hydrostatic force to the inward surface tension force: $\frac{1}{2} \rho g h^2 L = 2 T L$ $h^2 = \frac{4T}{\rho g}$ $h = \sqrt{\frac{4T}{\rho g}}$

4. Calculation: $h = \sqrt{\frac{4 \times 0.07}{10^3 \times 10}} = \sqrt{28 \times 10^{-6}} \text{ m} = \sqrt{28} \times 10^{-3} \text{ m} \approx 5.29 \text{ mm}$

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Final Answer:

Depending on the interpretation, the value of $h$ is:

• $3.74\text{ mm}$ (via Pressure Balance at the bottom)
• $5.29\text{ mm}$ (via Horizontal Force Balance)