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(C) The pressure inside the balloon due to surface tension is $P = \frac{2S}{R}$.Using Bernoulli's principle,the velocity of the gas escaping is $v =

Vedclass · Accepted Answer

$a=-\frac{1}{2}, \alpha=-\frac{1}{2}, \beta=-1, \gamma=\frac{1}{2}, \delta=\frac{7}{2}$

Vedclass · Answer

(C) The pressure inside the balloon due to surface tension is $P = \frac{2S}{R}$.Using Bernoulli's principle,the velocity of the gas escaping is $v = \sqrt{\frac{2P}{\rho}} = \sqrt{\frac{4S}{\rho R}} = 2S^{1/2} \rho^{-1/2} R^{-1/2}$.Given $\psi(r) \propto r^\alpha$,comparing with $v \propto R^{-1/2}$,we get $\alpha = -1/2$.The rate of change of volume is $\frac{dV}{dt} = -A \cdot v$.Since $V = \frac{4}{3}\pi R^3$,$\frac{dV}{dt} = 4\pi R^2 \frac{dR}{dt}$.So,$4\pi R^2 \frac{dR}{dt} = -A \cdot k \cdot S^{1/2} \rho^{-1/2} R^{-1/2}$,where $k$ is a constant.$R^{5/2} dR = -C \cdot S^{1/2} \rho^{-1/2} A dt$.Integrating from $R$ to $0$ and $0$ to $T$: $\int_0^R R^{5/2} dR = \int_0^T C' S^{1/2} \rho^{-1/2} A dt$.$\frac{2}{7} R^{7/2} = C' S^{1/2} \rho^{-1/2} A T$.Thus,$T \propto S^{-1/2} A^{-1} \rho^{1/2} R^{7/2}$.Comparing with $T \propto S^a A^\beta \rho^\gamma R^\delta$,we get $a = -1/2, \alpha = -1/2, \beta = -1, \gamma = 1/2, \delta = 7/2$.

Column $I$	Column $II$
$A$. Stoke's law	$I$. Pressure and energy
$B$. Turbulence	$II$. Hydraulic lift
$C$. Bernoulli's Principle	$III$. Viscous drag
$D$. Pascal's law	$IV$. Reynold's number

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Column $I$ Column $II$

$A$. Stoke's law $I$. Pressure and energy

$B$. Turbulence $II$. Hydraulic lift

$C$. Bernoulli's Principle $III$. Viscous drag

$D$. Pascal's law $IV$. Reynold's number

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Water flows through a frictionless duct with a cross-section varying as shown in the figure. Pressure $p$ at points along the axis is represented by:

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