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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:
A
$a = \frac{1}{2}, \alpha = \frac{1}{2}, \beta = -1, \gamma = +1, \delta = \frac{3}{2}$
B
$a = -\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = -\frac{1}{2}, \delta = \frac{5}{2}$
C
$a = -\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = \frac{1}{2}, \delta = \frac{7}{2}$
D
$a = \frac{1}{2}, \alpha = \frac{1}{2}, \beta = -\frac{1}{2}, \gamma = \frac{1}{2}, \delta = \frac{7}{2}$
Explanation
From Bernoulli\'s equation $v \propto \sqrt{\Delta P / \rho} = \sqrt{S / (\rho r)} \propto r^{-1/2} \implies a = -1/2$. Integrating volume discharge gives $T \propto S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$.
Detailed Solution
Excess pressure inside the balloon is $\Delta P = \frac{2S}{r}$. By Torricelli/Bernoulli principle, efflux velocity is $v(r) = \sqrt{\frac{2\Delta P}{\rho}} = \sqrt{\frac{4S}{\rho r}} \propto r^{-1/2}$, hence $a = -1/2$. The volumetric flow rate is $-\frac{dV}{dt} = A v(r) \implies -4\pi r^2 \frac{dr}{dt} = A \sqrt{\frac{4S}{\rho}} r^{-1/2} \implies dt = -\frac{4\pi}{2A} \sqrt{\frac{\rho}{S}} r^{5/2} dr$. Integrating from $R$ to 0 gives total deflation time $T = \frac{2\pi}{A} S^{-1/2} \rho^{1/2} \int_0^R r^{5/2} dr = \frac{2\pi}{A} S^{-1/2} \rho^{1/2} \left(\frac{2}{7} R^{7/2}\right) \propto S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$. Thus: $a = -1/2$, $\alpha = -1/2$, $\beta = -1$, $\gamma = 1/2$, $\delta = 7/2$.