Question

A balloon is made of a material of surface tension S and has a small outlet. It is filled with air of density \( \rho \). Initially the balloon is a sphere of radius R. When the gas is allowed to flow out slowly at a constant rate, its radius shrinks as \( r(t) \). Assume that the pressure inside the balloon is \( P(r) \) and is more than the outside pressure (\( P_0 \)) by an amount proportional to the surface tension and inversely proportional to the radius. The balloon bursts when its radius reaches \( r_0 \). Then the speed of gas coming out of the balloon at \( r = R \) is :

• A. \( \sqrt{\frac{S}{\rho R}} \)
• B. \( \sqrt{\frac{2S}{\rho R}} \)
• C. \( \sqrt{\frac{4S}{\rho R}} \)
• D. \( \sqrt{\frac{S}{2\rho R}} \)

Hint:

The excess pressure inside a spherical balloon due to surface tension is \( \Delta P = \frac{2S}{r} \). Apply Bernoulli's equation to relate the pressure difference to the speed of the escaping gas. Assume the initial speed of the gas inside the balloon is negligible.

Answer 1

The Correct Option is C

Step 1: Laplace Pressure Formula
Surface tension generates excess pressure within the balloon, described by the Laplace pressure formula: P(r) - P₀ = 2S / r. This formula indicates that the excess pressure is directly proportional to surface tension (S) and inversely proportional to the radius (r).

Step 2: Pressure Difference at Initial Radius
At the initial radius R, the pressure difference is calculated as: ΔP = P(R) - P₀ = 2S / R.

Step 3: Apply Bernoulli's Equation
Bernoulli's equation is used to determine the exit velocity of the gas. Assuming the gas velocity inside the balloon is negligible compared to the exit velocity and that the outlet is open to the atmosphere (pressure P₀), the equation simplifies to: P(R) + (1/2)ρ(0)² = P₀ + (1/2)ρv². This further simplifies to: P(R) - P₀ = (1/2)ρv².

Step 4: Calculate Gas Speed
By substituting the pressure difference expression: 2S / R = (1/2)ρv². Solving for v² yields: v² = (4S) / (ρR). Taking the square root gives the velocity: v = √(4S / (ρR)).

Conclusion:
The speed of the gas exiting the balloon at r = R is: v = √(4S / (ρR)).