A bubble having surface tension $T$ and radius $R$ is formed on a ring of radius $b$ ($b \ll R$). Air is blown inside the tube with velocity $v$ as shown. The air molecule collides perpendicularly with the wall of the bubble and stops. Calculate the radius at which the bubble separates from the ring.

Step 1: Identify the given parameters.

Let the air density be $\rho$, the surface tension of the bubble be $T$, the bubble radius be $R$, and the ring radius be $b$ where $b \ll R$.

Step 2: Calculate the force exerted by the air jet.

The air jet of cross-sectional area $\pi b^2$ enters the bubble with velocity $v$ and stops completely upon perpendicular impact with the far wall.

$$F_{\text{air}} = (\rho \pi b^2 v) v$$ $$F_{\text{air}} = \rho \pi b^2 v^2$$

Step 3: Determine the surface tension holding force.

The bubble film has two surfaces, giving a tension per unit length of $2T$ acting along the perimeter of the ring of radius $b$.

$$\sin\theta = \frac{b}{R}$$ $$F_{\text{ST}} = (2T)(2\pi b)\sin\theta$$ $$F_{\text{ST}} = \frac{4\pi T b^2}{R}$$

Step 4: Apply the condition for separation.

The bubble separates from the ring when the outward force from the air jet balances the inward holding force of the surface tension.

$$F_{\text{air}} = F_{\text{ST}}$$ $$\rho \pi b^2 v^2 = \frac{4\pi T b^2}{R}$$

Step 5: Solve for the separation radius $R$.

Simplify the equation by canceling common terms.

$$R = \frac{4T}{\rho v^2}$$

Answer: $R = \dfrac{4T}{\rho v^2}$