A glass capillary tube is of the shape of truncated cone with an apex angle α so that its two ends h

Q: Why is the angle of inclination of the wall taken as alpha/2 instead of alpha?

The apex angle of the cone is $\alpha$. The vertical axis of symmetry divides this angle into two equal halves. Therefore, each wall of the tube is inclined at an angle of $\alpha/2$ with respect to the vertical axis.

Q: What happens if the contact angle theta is zero?

If the contact angle $\theta = 0$, the formula simplifies to $h = \frac{2S}{b\rho g} \cos(\alpha/2)$. This represents the maximum possible capillary rise for this conical geometry.

Q: Does the height h depend on the radius at the bottom of the tube?

No, the height $h$ depends only on the radius of the tube $b$ at the exact level of the meniscus, because that is where the surface tension forces are acting to support the water column.

Q: What if the cone was inverted (narrower at the top)?

If the cone were inverted, the walls would tilt inwards instead of outwards. This would change the sign of the geometric correction term, resulting in an angle of $\theta - \alpha/2$ instead of $\theta + \alpha/2$.

Q: Is the meniscus shape perfectly spherical in a conical tube?

For very narrow capillary tubes, the meniscus can be approximated as a spherical cap. This approximation is highly accurate when the radius $b$ is much smaller than the capillary length of the liquid.

Question

A glass capillary tube is of the shape of truncated cone with an apex angle $\alpha$ so that its two ends have cross-sections of different radii. When dipped in water vertically, water rises in it to a height $h$, where the radius of its cross-section is $b$. If the surface tension of water is $S$, its density is $ho$, and its contact angle with glass is $	heta$, the value of $h$ will be ($g$ is the acceleration due to gravity)

Options:

(A) $\frac{2S}{bho g} \cos(	heta - \alpha)$

(B) $\frac{2S}{bho g} \cos(	heta + \alpha)$

(C) $\frac{2S}{bho g} \cos(	heta - \alpha/2)$

(D) $\frac{2S}{bho g} \cos(	heta + \alpha/2)$

Sigma Percentile · Accepted Answer

### Introduction to Capillary Action in Non-Cylindrical Tubes

Capillary action is one of the most fascinating phenomena in fluid mechanics.

We are all familiar with how water climbs up a narrow cylindrical straw.

But what happens when the geometry of the tube is modified?

In this problem, we explore a capillary tube shaped like a **truncated cone** with an apex angle $\alpha$.

This geometry introduces a beautiful interplay between fluid mechanics and trigonometry.

Let's dive deep into the physics and geometry of this setup!

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### The Physics of Equilibrium

When a capillary tube is dipped vertically into a liquid, the liquid rises until it reaches a stable height $h$.

At this height, the system is in a state of mechanical equilibrium.

This equilibrium is maintained by a balance of two competing pressures:

1. **Hydrostatic Pressure:** The weight of the raised liquid column exerts a downward pressure at the base, given by:
   $$P_{	ext{hydro}} = hho g$$

2. **Excess Pressure (Laplace Pressure):** The curved meniscus of the liquid creates a pressure difference across the interface. According to the Young-Laplace equation, this excess pressure is:
   $$\Delta P = \frac{2S}{R}$$
   where $R$ is the radius of curvature of the spherical meniscus.

Equating these two pressures gives us our master equation:
$$hho g = \frac{2S}{R}$$

To find the height $h$, our main task is to express the radius of curvature $R$ in terms of the physical dimensions of the tube.

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### The Geometric Challenge

In a standard cylindrical tube, the walls are vertical, and the radius of curvature $R$ is simply related to the tube radius $b$ by $R = \frac{b}{\cos	heta}$.

However, in our conical tube, the walls are tilted at an angle of $\frac{\alpha}{2}$ with respect to the vertical.

Let's analyze the angles at the contact line:

- The glass wall makes an angle of $\frac{\alpha}{2}$ with the vertical.
- The tangent to the liquid meniscus makes a contact angle $	heta$ with the glass wall.

By adding these angles, we find that the tangent to the meniscus makes an angle of $\left(	heta + \frac{\alpha}{2}ight)$ with the horizontal.

Since the radius of curvature $R$ is perpendicular to the meniscus tangent, it must make the same angle of $\left(	heta + \frac{\alpha}{2}ight)$ with the vertical axis of the tube.

Using the right-angled triangle formed by $R$ and the tube radius $b$ at height $h$, we get:
$$\cos\left(	heta + \frac{\alpha}{2}ight) = \frac{b}{R}$$

Rearranging this gives:
$$R = \frac{b}{\cos\left(	heta + \frac{\alpha}{2}ight)}$$

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### Final Calculation and Verification

Now, we substitute this expression for $R$ back into our master pressure balance equation:
$$hho g = \frac{2S}{\left(\frac{b}{\cos\left(	heta + \frac{\alpha}{2}ight)}ight)}$$

Simplifying this fraction yields the final expression for the height $h$:
$$h = \frac{2S}{bho g} \cos\left(	heta + \frac{\alpha}{2}ight)$$

This perfectly matches **Option (d)**!

To verify our result, let's look at the limiting case where $\alpha = 0$ (a perfect cylinder).

The formula reduces to:
$$h = \frac{2S\cos	heta}{bho g}$$

This is the classic Jurin's Law, confirming that our generalized formula is absolutely correct!

Select Answer:

Visualized Solution

Visualizing the Truncated Cone Capillary Tube

The Principle of Capillary Rise

Excess Pressure and Radius of Curvature

Analyzing the Geometry at the Contact Line

Finding the Angle of the Radius with the Horizontal

Relating $R$ and $b$

Substituting $R$ into the Pressure Balance

Solving for Height $h$

The Way Forward

The Sigma Insight: Surface Tension and Capillary Action

Introduction to Capillary Action in Non-Cylindrical Tubes

The Physics of Equilibrium

The Geometric Challenge

Final Calculation and Verification

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