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

The correct option is:
$a=-\frac{1}{2},\alpha =-\frac{1}{2},\beta =-1,\gamma =\frac{1}{2},\delta =\frac{7}{2}$

Step-by-Step Derivation:

1. Excess Pressure Inside the Balloon:
A spherical balloon made of a material with surface tension $S$ and radius $r$ has an excess pressure $\Delta P$ inside it given by:
$\Delta P=\frac{4S}{r}$

2. Velocity of the Outflow Gas:
Using Bernoulli's equation, the gauge pressure inside the balloon is converted into the kinetic energy of the escaping gas at the outlet. Thus, the velocity $v\left(r\right)$ of the gas of density $\rho$ flowing out is:
$v\left(r\right)=\sqrt{\frac{2\Delta P}{\rho }}=\sqrt{\frac{8S}{\rho r}}$
This shows that the speed of the gas depends on the radius $r$ as:
$v\left(r\right)\propto r^{-1/2}$
Comparing this with $v\left(r\right)\propto r^a$, we find:
$a=-\frac{1}{2}$

3. Equation of Continuity (Mass Flow Rate):
The rate of decrease in the volume of the balloon is equal to the rate at which gas volume leaves through the outlet of area $A$:
$-\frac{dV}{dt}=A·v\left(r\right)$
Since the volume of the spherical balloon is $V=\frac{4}{3}\pi r^3$, we have:
$-4\pi r^2\frac{dr}{dt}=A·\sqrt{\frac{8S}{\rho r}}$

4. Solving the Differential Equation:
Separating variables to find the total time $T$ for the radius to decrease from $R$ to $0$:
$-r^{5/2}dr=\frac{A}{4\pi }\sqrt{\frac{8S}{\rho }}dt$
Integrating both sides:
${\int }_0^R{r}^{5/2}dr=\frac{A\sqrt{8}{S}^{1/2}}{4\pi {\rho }^{1/2}}$
Evaluating the integral gives:
$\frac{2}{7}{R}^{7/2}=\frac{A\sqrt{8}{S}^{1/2}}{4\pi {\rho }^{1/2}}$
Rearranging for time $T$:
$T=\left(\frac{8\pi }{7\sqrt{8}}\right)S^{-1/2}{A}^{-1}{\rho }^{1/2}{R}^{7/2}$

5. Determining the Exponents:
This relationship matches the proportional form:
$T\propto S^{\alpha }{A}^{\beta }{\rho }^{\gamma }{R}^{\delta }$
By matching exponents, we get:
$\alpha =-\frac{1}{2},\beta =-1,\gamma =\frac{1}{2},\delta =\frac{7}{2}$