A balloon is made of a material of surface tension S and its inflation outlet (from where gas is filled in it) has small area A. It is filled with a gas of density ρ and takes a spherical shape of radius R. When the gas is allowed to flow freely out of it, its radius r changes from R to 0 (zero) in time T. If the speed v(r) of gas coming out of the balloon depends on r as  $r^a and T\propto S^{\alpha }{A}^{\beta }{\rho }^{\gamma }{R}^{\delta }$

The speed of the gas exiting the small outlet is assumed to follow

$v=r^a$

and the total discharge time obeys

$T\propto S^{\alpha }{A}^{\beta }{\rho }^{\gamma }{R}^{\delta }$

**1. Relating the velocity to the pressure difference** For a thin spherical membrane the Laplace law gives the excess internal pressure

$\mathrm{\Delta P}=\frac{2}{S}\frac{1}{r}$

Applying Bernoulli’s principle to the gas leaving the outlet, the characteristic speed is

$v\propto \sqrt{\frac{\mathrm{\Delta P}}{\rho }}=\sqrt{\frac{2}{S}\frac{1}{r}>·\sqrt{\frac{1}{\rho }>}}$

Thus

$v\propto S^{\frac{1}{2}}{\rho }^{\frac{-1}{2}}{r}^{\frac{-1}{2}}$

Comparing with the assumed form  gives the first exponent

$a=\frac{-}{1}2$

**2. Relating the discharge time to the parameters** The volume of the balloon is V = (4/3)πr³. The rate of volume loss through the outlet is the product of the outlet area A and the exit speed v:

$\frac{\mathrm{dV}}{\mathrm{dt}}=\; -A\; v$

Using V = (4/3)πr³, we have

$4\pi \; r²\; \backslash frac\{dr\}\{dt\}\; =\; -A\; v$

Insert the expression for v derived above (ignoring the constant 2 for scaling purposes):

$\backslash frac\{dr\}\{dt\}\; \propto \; -\backslash frac\{A\}\{4\pi \}\; S^\{1/2\}\; \rho ^\{-1/2\}\; r^\{a-2\}$

Since a = -1/2, the exponent on r is

$a-2\; =\; -\backslash frac\{1\}\{2\}\; -\; 2\; =\; -\backslash frac\{5\}\{2\}$

Thus

$\backslash frac\{dr\}\{dt\}\; \propto \; -A\; S^\{1/2\}\; \rho ^\{-1/2\}\; r^\{-5/2\}$

Separate variables and integrate from r = R at t = 0 to r = 0 at t = T:

$\backslash int\_\{R\}^\{0\}\; r^\{5/2\}\; dr\; \propto \; -A\; S^\{1/2\}\; \rho ^\{-1/2\}\; \backslash int\_\{0\}^\{T\}\; dt$

The left integral yields (2/7) R^{7/2}. Hence

$T\; \propto \; \backslash frac\{R^\{7/2\}\}\{A\; S^\{1/2\}\; \rho ^\{-1/2\}\}$

Re‑arranging gives the scaling

$T\; \propto \; S^\{-1/2\}\; A^\{-1\}\; \rho ^\{1/2\}\; R^\{7/2\}$

Comparing with the prescribed form  gives the remaining exponents

$\alpha \; =\; -\backslash frac\{1\}\{2\},\backslash ;\; \beta \; =\; -1,\backslash ;\; \gamma \; =\; \backslash frac\{1\}\{2\},\backslash ;\; \delta \; =\; \backslash frac\{7\}\{2\}$

**3. Summary of the exponents**

The dimensional‑analysis and physical‑law reasoning lead to the unique set

$a\; =\; -\backslash frac\{1\}\{2\},\backslash ;\; \alpha \; =\; -\backslash frac\{1\}\{2\},\backslash ;\; \beta \; =\; -1,\backslash ;\; \gamma \; =\; \backslash frac\{1\}\{2\},\backslash ;\; \delta \; =\; \backslash frac\{7\}\{2\}$

This matches the answer provided in the data.