### 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.