When water is filled carefully in a glass, one can fill it to a height $h$ above the rim of the glass due to the surface tension of water. To calculate $h$ just before water starts flowing, model the shape of the water above the rim as a disc of thickness $h$ having semicircular edges, as shown schematically in the figure. When the pressure of water at the bottom of this disc exceeds what can be withstood due to the surface tension, the water surface breaks near the rim and water starts flowing from there. If the density of water, its surface tension and the acceleration due to gravity are $10^{3}\,\text{kg m}^{-3}$, $0.07\,\text{Nm}^{-1}$ and $10\,\text{ms}^{-2}$, respectively, the value of $h$ (in mm) is ________.

Step 1: Identify the radius of curvature of the semicircular edge.

The water above the rim is modeled as a disc of thickness $h$ with semicircular edges, which gives the radius of curvature $r$.

$$r = \frac{h}{2}$$

Step 2: Determine the hydrostatic pressure at the bottom of the disc.

The gauge pressure at the bottom of the water disc of height $h$ is given by the hydrostatic pressure formula.

$$p_{\text{bottom}} = \rho g h$$

Step 3: Determine the pressure withstood by surface tension.

The excess pressure sustained by the curved surface of radius $r$ is given by Laplace's formula.

$$\Delta p = \frac{T}{r}$$

Substitute $r = \frac{h}{2}$ into the equation.

$$\Delta p = \frac{2T}{h}$$

Step 4: Set up the equilibrium condition.

Just before the water starts flowing, the hydrostatic pressure at the bottom equals the pressure withstood by the surface tension.

$$\rho g h = \frac{2T}{h}$$

Step 5: Solve for the height $h$.

Rearrange the equation to express $h$ in terms of the given parameters.

$$h^2 = \frac{2T}{\rho g}$$ $$h = \sqrt{\frac{2T}{\rho g}}$$

Step 6: Calculate the numerical value of $h$.

Substitute the given values: $\rho = 10^3\text{ kg m}^{-3}$, $T = 0.07\text{ N m}^{-1}$, and $g = 10\text{ m s}^{-2}$.

$$h = \sqrt{\frac{2 \times 0.07}{10^3 \times 10}}$$ $$h = \sqrt{1.4 \times 10^{-5}}$$ $$h \approx 3.74 \times 10^{-3}\text{ m}$$ $$h \approx 3.74\text{ mm}$$

Answer: $3.74$ (or $3.7$)