Animated Solution for Physics - Properties of Solids and Liquids: On heating water, bubbles being formed at the bottom of the vessel detatch and rise. Take the bubbles to be spheres of radius $R$ and making a circular contact of radius $r$ with the bottom of the vessel. If $r\ll R$ and the surface tension of water is $T$, value of $r$ just before bubbles detatch is (density of water is $\rho$)

Have you ever watched water boil and wondered about the physics behind those tiny bubbles forming at the bottom of the pan? It seems like a simple everyday event, but the moment a bubble decides to detach and rise to the surface is governed by a beautiful and delicate balance of forces. Let's dive into the mechanics of this phenomenon.

Buoyancy vs. Surface Tension

Imagine a vapor bubble of radius $R$ that has just formed at the bottom of a heated vessel. It is filled with hot vapor and is surrounded by denser liquid water. According to Archimedes' principle, the water exerts an upward buoyant force on the bubble. Since the bubble is essentially a sphere (with a tiny flat bottom), the buoyant force is given by the weight of the displaced water: where ${\rho }_w$ is the density of water and $g$ is the acceleration due to gravity.

If buoyancy is pushing it up, why doesn't it rise immediately? The answer lies in surface tension. The bubble is attached to the bottom of the vessel over a small circular contact area of radius $r$. The surface tension $T$ of the water acts along the perimeter of this contact circle, trying to minimize the surface area. It acts like a series of microscopic ropes pulling the bubble down, anchoring it to the surface.

To calculate the exact downward pull, we need to look at the direction of the surface tension force. The force acts tangentially to the bubble's surface at the contact line. If we draw a radius $R$ from the center of the bubble to the edge of the contact circle, it makes an angle $\theta$ with the vertical. Consequently, the tangent to the surface makes the same angle $\theta$ with the horizontal.

The total surface tension force acts along the circumference $2\pi r$. However, we only care about the vertical component that opposes buoyancy. This vertical component is $T\sin \theta$. Therefore, the total downward force is:

Here is where the problem gives us a crucial hint: $r\ll R$. This means the contact circle is very small compared to the bubble itself. If we look at the right-angled triangle formed by the bubble's center, the contact circle's center, and its edge, we can see that:

Substituting this geometric relation into our force equation, we get:

The bubble will detach at the exact moment the upward buoyant force overcomes the downward surface tension force. At this critical threshold, the two forces are equal:

Now, it's just a matter of careful algebra. Let's isolate $r^2$. First, we can cancel $\pi$ from both sides and divide by 2:

Multiplying both sides by $R$ and dividing by $T$, we get:

Finally, taking the square root gives us the critical contact radius just before detachment:

If you look at the options provided in the original JEE question, you will notice something surprising: none of them match our derived result! The correct expression contains a factor of $\sqrt{2/3}$, while the options have factors like $\sqrt{1/3}$, $\sqrt{1/6}$, etc.

This is a powerful lesson for any physics student. Sometimes, exam questions contain typographical errors in their options. When you have rigorously applied fundamental laws—like balancing buoyancy and surface tension—and carefully executed the geometry and algebra, you must trust your derivation. The physics is sound, the math is flawless, and the bubble will indeed detach when $r=R^2\sqrt{\frac{2{\rho }_{w}g}{3T}}$.