Animated Solution for Physics - Properties of Solids and Liquids: Two narrow bores of diameter $5.0\text{ mm}$ and $8.0\text{ mm}$ are joined together to form a U-shaped tube open at both ends. If this U-tube contains water, what is the difference in the level of two limbs of the tube. [Take surface tension of water $T=7.3\times 10^{-2}{\text{ Nm}}^{-1}$, angle of contact $=0$, $g=10{\text{ ms}}^{-2}$ and density of water $=1.0\times 10^3{\text{ kg m}}^{-3}$]

A Tale of Two Tubes

Imagine you are in a laboratory, and you have a U-shaped tube. But this isn't your standard, uniform U-tube. This one is a bit quirky—it's made by joining two narrow bores of different diameters. One side has a diameter of $5.0\text{ mm}$, making it quite thin, while the other side is slightly wider with a diameter of $8.0\text{ mm}$.

When you pour water into this tube, you might expect the water to settle at the exact same level on both sides, just like it does in a regular glass of water. But nature has a surprise for us! Because these tubes are so narrow, capillary action takes over. The water in the narrower tube climbs higher than the water in the wider tube. Our mission is to find exactly how much higher it climbs—the height difference, $h$.

Pressure and Meniscus

To solve this, we need to dive into the microscopic world of the water surface. At the top of each water column, the surface isn't flat; it curves upwards at the edges, forming a shape called a concave meniscus.

Because of surface tension, this curved surface acts like a stretched rubber membrane pulling upwards. This upward pull means that the pressure just below the surface of the water is actually slightly less than the atmospheric pressure pushing down from above.

This pressure drop is given by the formula $\frac{2T}{r}$, where $T$ is the surface tension and $r$ is the radius of the tube. Since the left tube is narrower (smaller $r$), the pressure drop is greater, which is exactly why the water has to rise higher on that side to compensate!

Equating and Simplifying

Now, let's use a powerful tool from fluid statics: In a continuous, static fluid, the pressure at any two points on the same horizontal level must be identical.

Let's pick a horizontal reference line near the bottom of the U-tube and mark two points, $A$ (in the left limb) and $B$ (in the right limb). We know that $p_A=p_B$.

Let's calculate the total pressure at point $A$ by starting from the top of the left limb and going down. We start with atmospheric pressure $p_0$, subtract the surface tension drop $\frac{2T}{r_A}$, and then add the pressure from the weight of the water column, which has a total height of $(h+y)$.

We do the exact same thing for point $B$ in the right limb. Here, the water column only has a height of $y$.

Since $p_A=p_B$, we can equate the two expressions:

Notice how beautifully this simplifies! The atmospheric pressure $p_0$ cancels out from both sides. When we expand the $\rho g(h+y)$ term, the $\rho gy$ also cancels out. We are left with a clean, elegant equation:

We are in the home stretch. It's time to plug in the numbers. But first, a crucial warning: always convert your units to standard SI units (meters, kilograms, seconds) to avoid silly mistakes!

The diameters are $5.0\text{ mm}$ and $8.0\text{ mm}$, so the radii are $r_A=2.5\times 10^{-3}\text{ m}$ and $r_B=4.0\times 10^{-3}\text{ m}$. The surface tension $T=7.3\times 10^{-2}\text{ N/m}$, density $\rho =10^3{\text{ kg/m}}^3$, and $g=10{\text{ m/s}}^2$.

Substituting these into our rearranged equation for $h$:

Let's factor out the $10^{-3}$ from the denominator of the fractions:

Converting this back to millimeters, we get our final answer: $h=2.19\text{ mm}$

The water in the narrower tube stands exactly $2.19\text{ mm}$ higher than in the wider tube. A perfect harmony of fluid mechanics and surface tension!