Q. 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 $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$
A) $a=\frac12,\alpha=\frac12,\beta=-1,\gamma=+1,\delta=\frac32$
B) $a=-\frac12,\alpha=-\frac12,\beta=-1,\gamma=-\frac12,\delta=\frac52$
C) $a=-\frac12,\alpha=-\frac12,\beta=-1,\gamma=\frac12,\delta=\frac72$
D) $a=\frac12,\alpha=\frac12,\beta=-\frac12,\gamma=\frac12,\delta=\frac72$
Explanation
[Solution by Claude - not from source] Excess pressure $\propto S/r$, so $v\propto\sqrt{S/(\rho r)}\Rightarrow a=-\frac12$. Flow rate: $Av\propto r^2\frac{dr}{dt}$, so $\frac{dr}{dt}\propto A\sqrt{S/\rho}\,r^{-5/2}$. Integrating from $R$ to 0: $T\propto\frac{\rho^{1/2}R^{7/2}}{S^{1/2}A}$. So $\alpha=-\frac12,\beta=-1,\gamma=\frac12,\delta=\frac72$. Answer (3).
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.