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

• A. $a=\frac{1}{2}, \alpha = \frac{1}{2}, \beta = \frac{1}{2}, \gamma = \frac{1}{2}, \delta = \frac{7}{2}$
• B. $a=\frac{1}{2}, \alpha = \frac{1}{2}, \beta = 1, \gamma = +1, \delta = \frac{3}{2}$
• C. $a=-\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = -\frac{1}{2}, \delta = \frac{5}{2}$
• D. $a=-\frac{1}{2}, \alpha = -\frac{1}{2}, \beta = -1, \gamma = \frac{1}{2}, \delta = \frac{7}{2}$

Answer 1

Answer: (D)

(4)

Answer 2

The excess pressure inside a spherical balloon of radius $r$ due to surface tension $S$ is given by $\Delta P = \frac{2S}{r}$.

The speed $v(r)$ of the gas flowing out of the balloon through an outlet of area $A$ is related to the pressure difference $\Delta P$ by Torricelli's law, $v = \sqrt{\frac{2 \Delta P}{\rho}}$, where $\rho$ is the density of the gas.

Substituting the expression for $\Delta P$, we get $v(r) = \sqrt{\frac{2(2S/r)}{\rho}} = \sqrt{\frac{4S}{\rho r}} = 2 \sqrt{\frac{S}{\rho}} r^{-1/2}$.

The problem states that $v(r) \propto r^a$. Comparing our expression $v(r) = \left(2 \sqrt{\frac{S}{\rho}}\right) r^{-1/2}$ with $v(r) \propto r^a$, we find $a = -1/2$.

The volume of the balloon is $V = \frac{4}{3} \pi r^3$. The rate of change of volume is $\frac{dV}{dt} = \frac{d}{dt}\left(\frac{4}{3} \pi r^3\right) = 4 \pi r^2 \frac{dr}{dt}$.

The rate at which gas flows out is the volume flow rate through the outlet, which is $A v(r)$. Since the gas is flowing out, the volume of the balloon is decreasing, so $\frac{dV}{dt} = - A v(r)$. $4 \pi r^2 \frac{dr}{dt} = - A \left(2 \sqrt{\frac{S}{\rho}} r^{-1/2}\right)$. $\frac{dr}{dt} = - \frac{2A \sqrt{S/\rho}}{4 \pi r^2} r^{-1/2} = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} r^{-5/2}$.

We need to find the time $T$ for the radius to change from $R$ to 0. We can integrate the differential equation: $\frac{dr}{dt} = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} r^{-5/2}$ $r^{5/2} dr = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} dt$.

Integrate from $t=0$ to $t=T$ and $r=R$ to $r=0$: $\int_R^0 r^{5/2} dr = \int_0^T - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} dt$. $\left[\frac{r^{5/2+1}}{5/2+1}\right]_R^0 = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} [t]_0^T$. $\left[\frac{r^{7/2}}{7/2}\right]_R^0 = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} T$. $\frac{2}{7} (0^{7/2} - R^{7/2}) = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} T$. $-\frac{2}{7} R^{7/2} = - \frac{A}{2 \pi} \sqrt{\frac{S}{\rho}} T$. $T = \frac{\frac{2}{7} R^{7/2}}{\frac{A}{2 \pi} \sqrt{\frac{S}{\rho}}} = \frac{2}{7} R^{7/2} \frac{2 \pi}{A} \sqrt{\frac{\rho}{S}} = \frac{4 \pi}{7} R^{7/2} A^{-1} \rho^{1/2} S^{-1/2}$.

The problem states $T \propto S^\alpha A^\beta \rho^\gamma R^\delta$. From our derived expression $T = \left(\frac{4 \pi}{7}\right) S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$, we can identify the exponents: $\alpha = -1/2$. $\beta = -1$. $\gamma = 1/2$. $\delta = 7/2$.

We found $a = -1/2$. So the exponents are $a = -1/2$, $\alpha = -1/2$, $\beta = -1$, $\gamma = 1/2$, $\delta = 7/2$. Comparing with the given options, option (4) matches these values.