A balloon is made of a material of surface tension $S$ and its inflation outlet (from where gas is f

Question

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

Professor PC Thomas Classes · Answer

$ \begin{array}{ll} 	ext {  }  	ext { (a) : } T \propto S^\alpha A^\beta ho^\gamma R^\delta \  {[T]=\left[\mathrm{MT}^{-2}ight]^\alpha\left[\mathrm{L}^2ight]^\beta\left[\mathrm{ML}^{-3}ight]^\gamma[\mathrm{L}]^\delta} \  \alpha+\gamma=0,2 \beta-3 \gamma+\delta=0,-2 \alpha=1 \ 	herefore \quad  \alpha=-1 / 2 ; \gamma=1 / 2 \end{array} $ \Now, $ \begin{aligned}  2 \beta-\frac{3}{2}+\delta \ & 2 \beta+\delta=\frac{3}{2} \end{aligned} $ \By checking with options $ \beta=-1, \delta=\frac{7}{2} 	ext { option (a) hold } $ So, $v \propto \sqrt{\Delta P} \propto \sqrt{\frac{s}{r}} \propto r^{-\frac{1}{2}}$ So, $a=-\frac{1}{2}$