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Q.21) 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) $a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, y=\frac{1}{2}, \delta=\frac{7}{2}$
Answers (1)
Solution:
Correct Option: (4)
The pressure inside the balloon due to surface tension is $P \propto \frac{S}{r}$, which drives the gas out. Equating this pressure energy with kinetic energy gives $v \propto r^{-1 / 2}$, so $a=-\frac{1}{2}$.
The total time is $T \propto \frac{\text { mass }}{\text { rate }} \propto \frac{\rho R^3}{\rho A v}=\frac{R^3}{A v}$. Substituting $v \propto \sqrt{\frac{S}{\rho R}}$, we get:
$$
T \propto \frac{R^3}{A} \cdot \sqrt{\frac{\rho R}{S}}=\frac{R^{7 / 2}}{A} \cdot \sqrt{\frac{\rho}{S}}
$$
So the exponents are:
$$
a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}
$$
Hence, the answer is option (4).
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