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(A) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius of the bubbl

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(A) The excess pressure inside a soap bubble is given by $\Delta P = \frac{4T}{r}$,where $T$ is the surface tension and $r$ is the radius of the bubble.
Since the pump increases the volume $V$ of the bubble at a constant rate,we have $\frac{dV}{dt} = k$ (constant).
Integrating this,$V = kt + C$. Assuming the bubble starts from zero volume at $t = 0$,$V = kt$.
Since $V = \frac{4}{3}\pi r^3$,we have $\frac{4}{3}\pi r^3 = kt$,which implies $r = \left(\frac{3kt}{4\pi}\right)^{1/3}$.
Substituting this into the expression for excess pressure: $\Delta P = \frac{4T}{r} = 4T \left(\frac{4\pi}{3kt}\right)^{1/3}$.
Thus,$\Delta P \propto t^{-1/3}$.
The pressure inside the bubble is $P = P_{atm} + \Delta P = P_{atm} + \frac{C'}{t^{1/3}}$,where $C'$ is a constant.
As $t$ increases,$P$ decreases. The graph representing this relationship is a curve that starts from a high value and decreases asymptotically towards $P_{atm}$ as $t$ increases,which corresponds to option $A$.

$A$ soap bubble is blown with the help of a mechanical pump at the mouth of a tube. The pump produces a constant increase per minute in the volume of the bubble,irrespective of its internal pressure. The graph between the pressure inside the soap bubble and time $t$ will be-

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