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 ra and T ∝ Sα Aβ ργ Rδ then

To find the relation for the velocity of the escaping gas and the total time taken for the balloon to deflate, we can use the concepts of surface tension, excess pressure, and fluid dynamics.

Step 1: Determine the excess pressure inside the balloon
A balloon of radius $r$ made of a material with surface tension $S$ has two surfaces (inner and outer). The excess pressure $\mathrm{\Delta P}$ inside the spherical balloon is given by:
$\mathrm{\Delta P}=\frac{4S}{r}$

Step 2: Find the speed of the escaping gas
Using Bernoulli's equation, the kinetic energy per unit volume of the gas leaving the outlet of area $A$ is equal to the excess pressure driving it out:
$\frac{1}{2}\rho v^2=\mathrm{\Delta P}$
Substituting $\mathrm{\Delta P}$ into the equation:
$\frac{1}{2}\rho v^2=\frac{4S}{r}$
Solving for the speed of the gas $v$:
$v=\sqrt{\frac{8S}{\rho r}}={\left(\frac{8S}{\rho }\right)}^{\frac{1}{2}}{r}^{-\frac{1}{2}}$
Since $v\left(r\right)\propto r^a$, we find:
$a=-\frac{1}{2}$

Step 3: Relate the rate of deflation to the volume flow rate
The rate at which the volume of the balloon decreases must equal the rate at which gas flows out through the outlet of area $A$:
$-\frac{dV}{dt}=Av$
Since the volume of a sphere is $V=\frac{4}{3}\pi r^3$, we have $dV=4\pi r^{2}dr$:
$-4\pi r^2\frac{dr}{dt}=A{\left(\frac{8S}{\rho }\right)}^{\frac{1}{2}}{r}^{-\frac{1}{2}}$
Rearranging the terms to integrate:
$-r^{\frac{5}{2}}dr=\frac{A}{4\pi }{\left(\frac{8S}{\rho }\right)}^{\frac{1}{2}}dt$

Step 4: Integrate to find the total time T
Integrating the radius from $R$ to $0$ over the time interval from $0$ to $T$:
${\int }_R^0-r^{\frac{5}{2}}dr=\frac{A}{4\pi }{\left(\frac{8S}{\rho }\right)}^{\frac{1}{2}}{\int }_0^{T}dt$
${\left[\frac{2}{7}{r}^{\frac{7}{2}}\right]}_0^R=\frac{A}{4\pi }{\left(\frac{8S}{\rho }\right)}^{\frac{1}{2}}T$
$\frac{2}{7}{R}^{\frac{7}{2}}=\frac{A}{4\pi }{\left(\frac{8S}{\rho }\right)}^{\frac{1}{2}}T$
Solving for $T$:
$T=\frac{8\pi }{7A}{\left(\frac{\rho }{8S}\right)}^{\frac{1}{2}}{R}^{\frac{7}{2}}$
Expressing this in terms of proportionality:
$T\propto S^{-\frac{1}{2}}{A}^{-1}{\rho }^{\frac{1}{2}}{R}^{\frac{7}{2}}$

Comparing this to the given relation $T\propto S^{\alpha }{A}^{\beta }{\rho }^{\gamma }{R}^{\delta }$, we obtain:
$\alpha =-\frac{1}{2},\beta =-1,\gamma =\frac{1}{2},\delta =\frac{7}{2}$