Step-by-step Solution

1. Understanding the Physical Principle The problem describes water filled in a glass to a height h above the rim. This is possible due to surface tension, which creates a pressure difference across the curved surface of the water, holding the water bulge in place. Water starts to flow when the hydrostatic pressure exerted by the column of water of height h exceeds the excess pressure that can be sustained by the surface tension.

2. Formulating the Pressure Balance Equation The hydrostatic gauge pressure at the bottom of the water disc (at the level of the rim) is given by: $P_{hydrostatic} = \rho g h$ where $\rho$ is the density of water, g is the acceleration due to gravity, and h is the height of the water above the rim.

   The excess pressure inside the curved liquid surface is given by the Young-Laplace equation: $\Delta P = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$ where T is the surface tension, and $R_1$ and $R_2$ are the principal radii of curvature of the surface.

   At the point of breaking, the hydrostatic pressure is equal to the maximum excess pressure the surface tension can provide: $P_{hydrostatic} = \Delta P$ $\rho g h = T \left( \frac{1}{R_1} + \frac{1}{R_2} \right)$

3. Determining the Radii of Curvature from the Model The problem specifies a model for the water shape: "a disc of thickness h having semicircular edges".

   • $R_1$: This is the radius of curvature in the vertical plane (the cross-section shown in the figure). The edge is described as semicircular with the thickness h as its diameter. Therefore, the radius of this semicircle is $R_1 = h/2$.
   • $R_2$: This is the radius of curvature in the horizontal plane. For a circular glass, this would be the radius of the glass rim. However, the radius of the glass is not provided. In such problems, it's common to assume that the radius of the glass is much larger than the height h ($R_{glass} \gg h$), or to model the situation as a 2D problem (like water in a long trough). In either case, the curvature in the horizontal plane is considered negligible, which means $R_2 \rightarrow \infty$ and $1/R_2 \rightarrow 0$.

4. Substituting Radii and Solving for h Substituting the values of $R_1$ and $R_2$ into the pressure balance equation: $\rho g h = T \left( \frac{1}{h/2} + 0 \right)$ $\rho g h = \frac{2T}{h}$ Now, we solve for h: $\rho g h^2 = 2T$ $h^2 = \frac{2T}{\rho g}$ $h = \sqrt{\frac{2T}{\rho g}}$

5. Calculation with Given Values We are given:

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

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

   To find the numerical value, we calculate $\sqrt{14}$: $\sqrt{14} \approx 3.74165...$ So, $h \approx 3.74 \times 10^{-3} \text{ m}$.

6. Converting to Millimeters The question asks for the value of h in millimeters (mm). To convert from meters to millimeters, we multiply by 1000: $h = (3.74 \times 10^{-3}) \times 1000 \text{ mm}$ $h = 3.74 \text{ mm}$

   The value of h (in mm) is 3.74.