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 $\rho$ 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 $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$ then
Answer: (C) $a = -\frac{1}{2},\ \alpha = -\frac{1}{2},\ \beta = -1,\ \gamma = \frac{1}{2},\ \delta = \frac{7}{2}$
Excess pressure inside the balloon: $\Delta P = \dfrac{kS}{r}$ ($k$ is a number, $2$ or $4$).
Bernoulli at the outlet: $\dfrac{1}{2}\rho v^2 = \dfrac{kS}{r}$, so
$$v = \sqrt{\frac{2kS}{\rho r}} \propto r^{-1/2} \;\Rightarrow\; a = -\frac{1}{2}$$
Rate of volume loss:
$$-\frac{d}{dt}\left(\frac{4}{3}\pi r^3\right) = Av \;\Rightarrow\; -4\pi r^2\frac{dr}{dt} = A\sqrt{\frac{2kS}{\rho}}\,r^{-1/2}$$
$$dt = -\frac{4\pi}{A}\sqrt{\frac{\rho}{2kS}}\,r^{5/2}\,dr \;\Rightarrow\; T = \frac{4\pi}{A}\sqrt{\frac{\rho}{2kS}}\cdot\frac{2}{7}R^{7/2}$$
So $T \propto S^{-1/2}A^{-1}\rho^{1/2}R^{7/2}$: $\alpha = -\frac{1}{2}$, $\beta = -1$, $\gamma = \frac{1}{2}$, $\delta = \frac{7}{2}$.
Solution by Sreeraj P, M.Sc Physics