JEE Main 12 Jan 2019 Evening Question 7
Question: A soap bubble, blown by a mechanical pump at the mouth of a tube, increases in volume, with time, at a constant rate. The graph that correctly depicts the time dependence of pressure inside the bubble is given by [JEE Main Online Paper Held On 12-Jan-2019 Evening]
Options:
A)
B)
C)
D)
E) None of these
Show Answer
Answer:
Correct Answer: E
Solution:
$ \frac{dV}{dt}=\cos nt. $ or, $ \frac{d( \frac{4}{3}\pi r^{3} )}{dt}=K $ $ \frac{4}{3}\pi (3r^{2})\frac{dr}{dt}=K,4\pi r^{2}\frac{dr}{dt}=K $
$ \Rightarrow $ $ r^{3}=Kt+C\Rightarrow r=K _1{t^{1/3}}+C _1 $
The excess pressure inside the bubble is $ \frac{4S}{r} $
i.e. $ P _{excess}\propto \frac{1}{r} $ pressure, $ P=P _0+\frac{4S}{r} $ $ =P _0+\frac{4S}{K _1{t^{1/3}}+C _1} $
None of the given options is correct.
