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) $a = -\frac{1}{2},\ \alpha = -\frac{1}{2},\ \beta = -1,\ \gamma = \frac{1}{2},\ \delta = \frac{7}{2}$
Concept. The excess pressure $\dfrac{kS}{r}$ drives the gas out (Bernoulli); the volume loss rate is $Av$.
Step 1. Exit speed. $\dfrac12\rho v^2 = \dfrac{kS}{r} \Rightarrow v \propto r^{-1/2}$, so $a = -\dfrac12$.
Step 2. Volume loss.
$$-4\pi r^2\frac{dr}{dt} = A\sqrt{\frac{2kS}{\rho}}\,r^{-1/2}$$
Step 3. Integrate.
$$T = \frac{4\pi}{A}\sqrt{\frac{\rho}{2kS}} \cdot \frac27R^{7/2} \;\Rightarrow\; T \propto S^{-1/2}A^{-1}\rho^{1/2}R^{7/2}$$
Answer. $a = -\dfrac12,\ \alpha = -\dfrac12,\ \beta = -1,\ \gamma = \dfrac12,\ \delta = \dfrac72$ (Option C).
Solution by Sreeraj P, M.Sc Physics