Properties Of Matter Ques 12
12 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
(2019 Main, 12 Jan II)
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Answer:
Correct Answer: 12.( b )
Solution:
2 When soap bubble is being inflated and its temperature remains constant, then it follows Boyle’s law, so
$ \begin{array}{rl} & p V=\operatorname{constant}(k) \\ \Rightarrow p & p \frac{k}{V} \end{array} $
Differentiating above equation with time, we get
$ \frac{d p}{d t}=k \cdot \frac{d}{d t}\left(\frac{1}{V}\right) \Rightarrow \frac{d p}{d t}=k\left(\frac{-1}{V^2}\right) \cdot \frac{d V}{d t} $
It is given that, $\frac{d V}{d t}=c$ (a constant) So, $ \frac{d p}{d t}=\frac{-k c}{V^2} $
Now, from $\frac{d V}{d t}=c$; we get
or
$ \begin{aligned} & d V=c d t \\ & \int d V=\int c d t \text { or } V=c t \end{aligned} $
From Eqs. (i) and (ii), we get
$ \frac{d p}{d t}=\frac{-k c}{c^2 t^2} \text { or } \frac{d p}{d t}=-\left(\frac{k}{c}\right) t^{-2} \Rightarrow d p=-\frac{k}{c} \cdot t^{-2} d t $
Integrating both sides, we get
$ \begin{aligned} \int d p & =-\frac{k}{c} \int t^{-2} d t \\ p & =-\frac{k}{c} \cdot\left(\frac{t^{-2+1}}{-2+1}\right) \\ & =-\frac{k}{c} \cdot \frac{-1}{t}=\frac{k}{c t} \text { or } p \propto \frac{1}{t} \end{aligned} $
Hence, $p$ versus $\frac{1}{t}$ graph is a straight line, which is correctly represented in option (b).
BETA
