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 $$10^{3}kg m^{−3}, 0.07 Nm^{−1}$$ and $$10 ms^{−2}$$, respectively, the value of h (in mm) is _________.

Let the radius of the rim of the glass be $$R$$ and let the thickness of the bulging water column above the rim be $$h$$.

The free surface of the water near the rim is approximated as a semicylinder whose cross-section is a semicircle of radius $$h/2$$. According to Laplace’s law, the excess pressure that such a curved surface can sustain is

$$\Delta P = T\left(\frac1{R_1} + \frac1{R_2}\right),$$

where $$R_1$$ and $$R_2$$ are the two principal radii of curvature.
Here, $$R_1 = R$$ (large radius along the circular rim) and $$R_2 = h/2$$ (radius of the semicircle in the vertical plane). Thus

$$\Delta P = T\left(\frac1{R} + \frac{2}{h}\right).$$

A normal drinking glass has $$R \gg h$$, so $$1/R$$ is negligible in comparison with $$2/h$$. Hence we set

$$\Delta P \approx \frac{2T}{h}.$$

The hydrostatic gauge pressure at the bottom of the bulging column is

$$P_{\text{hydro}} = \rho g h.$$

The water starts to spill when this hydrostatic pressure just equals the maximum pressure that surface tension can withstand, i.e.

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

Solving for $$h$$:

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

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

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

Therefore

$$h \approx 3.7\,\text{mm} \;\;(\text{rounded to one significant figure, } h \simeq 4\,\text{mm}).$$

Hence the height to which water can be filled above the rim is about 4 mm.