Question 24. 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^{a} and T \propto S^{\alpha} A^{\beta} \rho^{\gamma} R^{\delta}
(1) a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-1, \gamma=+1, \delta=\frac{3}{2}
(2) a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=-\frac{1}{2}, \delta=\frac{5}{2}
(3) a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}
(4) a=\frac{1}{2}, \alpha=\frac{1}{2}, \beta=-\frac{1}{2}, \gamma=\frac{1}{2}, \delta=\frac{7}{2}
Answer: Option (3)
Explanation:
1. Pressure inside a spherical balloon due to surface tension (Laplace’s law) is
\Delta P=\dfrac{2S}{r}.
This pressure difference drives the outflow.
2. Using Bernoulli (dynamic pressure ~ \tfrac{1}{2}\rho v^{2})
for the gas leaving the outlet, the outflow speed scales as
v(r)\sim\sqrt{\dfrac{2\Delta P}{\rho}}=\sqrt{\dfrac{2(2S/r)}{\rho}}=\sqrt{\dfrac{4S}{\rho}}\;r^{-1/2}.So the power-law exponent is
a=-\dfrac{1}{2}.3. Volume of the spherical balloon is
V=\dfrac{4}{3}\pi r^{3}.
The rate of change of volume equals the outflow through area A:
\dfrac{dV}{dt}=-A\,v(r).Writing \dfrac{dV}{dr}=4\pi r^{2}
we get
4\pi r^{2}\dfrac{dr}{dt}=-A\,v(r).Substitute v(r)=k\,r^{a} (with k\propto S^{1/2}\rho^{-1/2}):
\dfrac{dr}{dt}=-\dfrac{A k}{4\pi}\;r^{\,a-2}.Separate variables and integrate from r=R to 0 to find time T:
T=\dfrac{4\pi}{A k}\int_{0}^{R} r^{\,2-a}\,dr=\dfrac{4\pi}{A k}\cdot\dfrac{R^{\,3-a}}{3-a}.\!Thus up to a numerical factor T\propto\dfrac{1}{A k}\;R^{\,3-a}.
4. From step 2, k\propto\sqrt{\dfrac{S}{\rho}}=S^{1/2}\rho^{-1/2}.
Substitute into the expression for T:
T\propto \dfrac{1}{A}\;S^{-1/2}\rho^{1/2}\;R^{\,3-a}.Comparing with T\propto S^{\alpha}A^{\beta}\rho^{\gamma}R^{\delta} gives
\alpha=-\tfrac{1}{2},\quad \beta=-1,\quad \gamma=+\tfrac{1}{2},\quad \delta=3-a.Using a=-\tfrac{1}{2},
we get \delta=3-(-\tfrac{1}{2})=\tfrac{7}{2}.
Therefore the set of exponents is
a=-\tfrac{1}{2},\ \alpha=-\tfrac{1}{2},\ \beta=-1,\ \gamma=\tfrac{1}{2},\ \delta=\tfrac{7}{2},which matches Option (3).