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
The excess pressure inside the balloon due to surface tension is given by the Laplace pressure formula: P(r) - P₀ = 2S / r. This shows that the excess pressure is proportional to surface tension (S) and inversely proportional to the radius (r), which matches the Laplace pressure formula.

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

Step 3: Apply Bernoulli's Equation
We can use Bernoulli's equation to find the speed of the gas coming out of the outlet. Assuming the speed of the gas inside the balloon is negligible compared to the speed of the gas coming out, and assuming the outlet is open to the atmosphere (pressure P₀), we can write Bernoulli's equation as: P(R) + (1/2)ρ(0)² = P₀ + (1/2)ρv².
Simplifying: P(R) - P₀ = (1/2)ρv².

Step 4: Solve for the Speed of the Gas
Substituting the expression for the pressure difference: 2S / R = (1/2)ρv².
Solving for v²: v² = (4S) / (ρR).
Taking the square root of both sides: v = √(4S / (ρR)).

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