When water is filled carefully in a glass, one can fill it to a height h above the rim of the glass

Q: Why do we ignore the radius of the glass in the calculation?

The Young-Laplace equation uses the sum of the inverses of the radii ($1/R_1 + 1/R_2$). Because the radius of the glass ($R_1$) is much larger than the tiny radius of the semicircular edge ($R_2$), the term $1/R_1$ is extremely small compared to $1/R_2$. Ignoring it simplifies the math without significantly affecting the accuracy of the result.

Q: How do we know the radius of the edge is $h/2$?

The problem explicitly instructs us to model the edge as 'semicircular' and states the total thickness of the disc is $h$. The thickness of a semicircle represents its diameter. Therefore, the radius is half of the diameter, which is $h/2$.

Q: What is the pressure outside the water bulge?

The pressure immediately outside the water bulge is the standard atmospheric pressure, often denoted as $P_0$.

Q: Why do we equate $\rho g h$ to the excess pressure $2T/h$?

The water column creates an outward hydrostatic pressure of $\rho g h$ at its base. The surface tension creates an inward 'squeezing' pressure of $2T/h$. At the exact moment before the water spills, these two opposing pressures are perfectly balanced.

Q: What physically happens if we add water beyond the height $h$?

If you add more water, the hydrostatic pressure $\rho g h$ increases. However, the surface tension cannot provide any more inward pressure to balance it. The 'skin' of the water breaks, the equilibrium is lost, and the water spills over the rim.

Question

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

Sigma Percentile · Accepted Answer

Have you ever poured water into a glass right up to the brim, and then carefully added just a few more drops? You'll notice that the water doesn't spill immediately. Instead, it forms a beautiful, convex bulge above the rim. This phenomenon is a classic demonstration of **surface tension** acting like an invisible, elastic skin holding the water together.

In this problem, we are asked to find the maximum height $h$ this bulge can reach before the "skin" breaks and the water spills. The problem simplifies the geometry for us: we model the water above the rim as a disc of thickness $h$ with semicircular edges.

### Analyzing the Geometry of the Bulge

Let's focus on the edge of this water disc. Since the total thickness of the disc is $h$, and the edge is semicircular, the diameter of this semicircle must be $h$. Consequently, the radius of curvature of this edge is $r = \frac{h}{2}$.

Now, we must recall the **Young-Laplace equation**, which governs the pressure difference across any curved liquid interface. The equation states that the excess pressure $\Delta P$ is given by:

$$\Delta P = T \left(\frac{1}{R_1} + \frac{1}{R_2}ight)$$

Here, $T$ is the surface tension, and $R_1$ and $R_2$ are the two principal radii of curvature of the surface.

### The Crucial Approximation

What are $R_1$ and $R_2$ in our case? One radius of curvature, let's call it $R_2$, is the radius of the semicircular edge we just found: $R_2 = \frac{h}{2}$. The other radius of curvature, $R_1$, corresponds to the macroscopic curvature of the glass itself (the radius of the disc).

Here is where physical intuition comes into play. A typical glass has a radius of a few centimeters, while the bulge height $h$ is only a few millimeters. Therefore, $R_1 \gg R_2$. Because $R_1$ is so much larger, its inverse $\frac{1}{R_1}$ is negligibly small compared to $\frac{1}{R_2}$. We can safely approximate $\frac{1}{R_1} \approx 0$.

This simplifies our Young-Laplace equation dramatically:

$$\Delta P \approx T \left(0 + \frac{1}{h/2}ight) = \frac{2T}{h}$$

This $\frac{2T}{h}$ is the maximum capillary pressure the surface tension can provide to hold the water in.

### Balancing the Forces

What is trying to push the water out? It is the hydrostatic pressure created by the weight of the water column itself. At the bottom of the bulge (right at the rim of the glass), the hydrostatic pressure is:

$$P_{hydro} = ho g h$$

Just before the water starts flowing, the system is at the absolute limit of its stability. The outward hydrostatic pressure is perfectly balanced by the inward capillary pressure. Equating the two gives us our master equation:

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

### Final Calculation

Now, it is a straightforward algebraic exercise. Rearranging the equation to solve for $h$, we get:

$$h^2 = \frac{2T}{ho g}$$

We are given the following values:
- Surface tension, $T = 0.07 	ext{ Nm}^{-1}$
- Density of water, $ho = 10^3 	ext{ kg m}^{-3}$
- Acceleration due to gravity, $g = 10 	ext{ ms}^{-2}$

Substituting these values into our equation:

$$h^2 = \frac{2 	imes 0.07}{10^3 	imes 10} = \frac{0.14}{10^4} = 14 	imes 10^{-6} 	ext{ m}^2$$

Taking the square root of both sides:

$$h = \sqrt{14 	imes 10^{-6}} 	ext{ m} = \sqrt{14} 	imes 10^{-3} 	ext{ m}$$

Since $10^{-3} 	ext{ m}$ is exactly $1 	ext{ mm}$, we have:

$$h = \sqrt{14} 	ext{ mm} \approx 3.741 	ext{ mm}$$

Thus, the water can bulge to a maximum height of approximately **$3.74 	ext{ mm}$** before it breaks the surface tension and spills over.

Enter Numerical Value:

Visualized Solution

Visualizing the Water Bulge

Radius of the Semicircular Edge

Young-Laplace Equation

Approximating the Radii

Capillary Pressure

Hydrostatic Pressure

Balancing the Pressures

Rearranging for Height

Substituting Values

Final Calculation

The Sigma Insight: Surface Tension and Capillary Action

Analyzing the Geometry of the Bulge

The Crucial Approximation

Balancing the Forces

Final Calculation

Similar Questions

A small spherical droplet of density $d$ is floating exactly half immersed in a liquid of density $ρ$ and surface tension $T$ . The radius of the droplet is (take note that the surface tension applies an upward force on the droplet)