A soap bubble is being blown at the end of very narrow tube of radius b. Air (density rho) moves wit

Q: Why did we use $2T$ instead of $T$ for the surface tension force?

A soap bubble is a thin liquid film suspended in air, meaning it has two free surfaces in contact with air—one on the inside of the bubble and one on the outside. Since surface tension acts on every free surface, the effective force per unit length is doubled, making it $2T$.

Q: How does the air coming to rest exert a force?

The air entering the tube has a velocity $v$ and thus carries momentum. When it enters the bubble, it slows down and comes to rest. By Newton's second law, this change in momentum over time exerts a continuous thrust force on the bubble, similar to a rocket engine or water spraying from a hose.

Q: Why is the final radius independent of the tube's radius $b$?

While a wider tube increases the mass flow rate of air (which increases the pushing force by $b^2$), it also increases the circumference of contact and thus the holding force of surface tension (which also scales with $b^2$). Because both forces scale with $b^2$, the tube radius cancels out completely at the moment of detachment.

Q: What happens if we blow air faster?

Since the critical radius is inversely proportional to the square of the velocity ($r \propto 1/v^2$), blowing air faster dramatically decreases the size at which the bubble detaches. To blow large bubbles, you must keep the velocity $v$ as small as possible.

Q: Is this formula valid for a water droplet instead of a soap bubble?

For a liquid droplet (like water dripping from a tap), there is only one free surface (the outer one). In that case, the surface tension force per unit length would be $T$ instead of $2T$, leading to a critical radius of $r = \frac{2T}{\rho v^2}$.

Question

A soap bubble is being blown at the end of very narrow tube of radius $b$. Air (density $\rho$) moves with a velocity $v$ inside the tube and comes to rest inside the bubble. The surface tension of the soap solution is $T$. After sometime the bubble, having grown to radius $r$ separates from the tube. Find the value of $r$. Assume that $r \gg b$ so, that you can consider the air to be falling normally on the bubble's surface.

Sigma Percentile · Accepted Answer

### The Physics of Bubble Detachment

Imagine blowing a delicate soap bubble at the end of a thin straw. As you blow air into it, the bubble expands, shimmering with iridescent colors, until suddenly—it detaches and floats away. Have you ever wondered what physical laws govern the exact moment of this detachment?

This classic problem from the **JEE Advanced 2003** exam invites us to explore the beautiful interplay between **fluid dynamics (momentum transfer)** and **surface chemistry (surface tension)**.

Let's dive deep into the mechanics of this phenomenon.

---

### The Battle of Forces

At the mouth of the tube, a silent tug-of-war is taking place. On one side, we have the incoming stream of air pushing the bubble outward. On the other side, the cohesive forces of surface tension are desperately trying to hold the bubble anchored to the rim of the tube.

To find the critical size at which the bubble detaches, we must analyze these two competing forces at the threshold of equilibrium.

---

### 1. The Outward Thrust of Air Flow

When air of density $ho$ flows through a tube of radius $b$ with a velocity $v$, it carries momentum. As this air enters the bubble, it spreads out and comes to rest relative to the bubble's expanding boundary.

According to **Newton's Second Law of Motion**, the force exerted on a system is equal to the rate of change of its momentum.

First, let's find the mass of air entering the bubble per second (mass flow rate):

$$\frac{dm}{dt} = ho \cdot A \cdot v = ho (\pi b^2) v$$

Since the air comes to rest inside the bubble, its change in velocity is $\Delta v = v - 0 = v$.

Therefore, the rate of change of momentum—which is the outward force exerted by the air—is:

$$F_{	ext{air}} = \frac{dp}{dt} = \left(\frac{dm}{dt}ight) v = ho (\pi b^2) v^2$$

This force acts vertically downwards, tending to push the bubble off the tube.

---

### 2. The Holding Force of Surface Tension

Now, let's look at the force holding the bubble. Surface tension $T$ acts along the line of contact between the liquid film and the tube's circular rim of radius $b$.

Here is a crucial detail: **a soap bubble has two free surfaces** (an inner surface and an outer surface). Therefore, the effective surface tension force per unit length is $2T$.

The total length of contact is the circumference of the tube rim:

$$L = 2\pi b$$

Thus, the total surface tension force acting along the tangent to the bubble surface is:

$$F_{	ext{total}} = (2T) \cdot (2\pi b) = 4\pi b T$$

However, this force acts along the tangent to the bubble's surface at the contact point. To find the force holding the bubble vertically upward against the air flow, we must resolve this force into its vertical component:

$$F_{	ext{ST}} = F_{	ext{total}} \sin	heta = 4\pi b T \sin	heta$$

where $	heta$ is the angle that the radius vector to the contact point makes with the vertical axis.

---

### 3. The Geometry of the Bubble

From the geometry of the spherical bubble of radius $r$ attached to the tube of radius $b$, we can construct a right-angled triangle where:
- The hypotenuse is the radius of the bubble, $r$.
- The side opposite to the angle $	heta$ is the radius of the tube, $b$.

From this triangle, we get a simple trigonometric relation:

$$\sin	heta = \frac{b}{r}$$

Substituting this back into our vertical surface tension force equation:

$$F_{	ext{ST}} = 4\pi b T \left(\frac{b}{r}ight) = \frac{4\pi b^2 T}{r}$$

Notice the beautiful physical insight here: as the bubble grows larger ($r$ increases), the angle $	heta$ decreases, making the tangent more horizontal. Consequently, the vertical component of the surface tension force holding the bubble decreases as $1/r$.

---

### 4. The Threshold of Detachment

The bubble will detach at the exact instant when the outward force of the air flow overcomes the vertical holding force of surface tension:

$$F_{	ext{air}} = F_{	ext{ST}}$$

$$ho (\pi b^2) v^2 = \frac{4\pi b^2 T}{r}$$

We can beautifully cancel the common terms $\pi$ and $b^2$ from both sides of the equation:

$$ho v^2 = \frac{4T}{r}$$

Solving for the critical radius $r$:

$$r = \frac{4T}{ho v^2}$$

### Conclusion

This elegant result shows that the critical radius of the bubble is **independent of the tube's radius $b$**! It depends only on the surface tension of the liquid $T$, the density of the air $ho$, and the square of the blowing velocity $v^2$.

If you blow harder (increase $v$), the bubble will detach at a much smaller radius $r$. To get large bubbles, you must blow very gently!

Visualized Solution

Visualizing the Setup

Understanding the Force Exerted by Air Flow

Calculating the Thrust Force

Analyzing the Surface Tension Force

Resolving the Surface Tension Force

Geometric Relation for $sin θ$

Substituting $sin θ$ into the Force Equation

Equilibrium at the Moment of Detachment

Solving for the Critical Radius $r$

The Sigma Insight: Surface Tension and Capillary Action

The Physics of Bubble Detachment

The Battle of Forces

1

2

3

4

Conclusion

Similar Questions

A glass tube of uniform internal radius ( $r$ ) has a valve separating the two identical ends. Initially, the valve is in a tightly closed position. End 1 has a hemispherical soap bubble of radius $r$ . End 2 has sub-hemispherical soap bubble as shown in figure. Just after opening the valve.

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)

Two spherical soap bubbles of radii $r_{1}$ and $r_{2}$ in vacuum combine under isothermal conditions. The resulting bubble has a radius equal to

A soap bubble of radius 3 cm is formed inside the another soap bubble of radius 6 cm. The radius of an equivalent soap bubble which has the same excess pressure as inside the smaller bubble with respect to the atmospheric pressure is ........ cm.

When two soap bubbles of radii $a$ and $b (b > a)$ coalesce, the radius of curvature of common surface is

Work done in increasing the size of a soap bubble from radius of $3 cm$ to $5 cm$ is nearly (surface tension of soap solution = $0.03 Nm^{- 1}$ )