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When soap bubble is being inflated and its temperature remains constant, then it follows Boyle's law, so
pV = constant (k)
$ \Rightarrow p = {k \over V}$
Differentiating above equation with time, we get
${{dp} \over {dt}} = k.\,{d \over {dt}}\left( {{1 \over V}} \right)$
$ \Rightarrow {{dp} \over {dt}} = k\left( {{{ - 1} \over {{V^2}}}} \right).\,{{dV} \over {dt}}$
It is given that, ${{dV} \over {dt}} = c$ (a constant)
So, ${{dp} \over {dt}} = {{ - kc} \over {{V^2}}}$ ..... (i)
Now, from ${{dV} \over {dt}} = c$, we get
$dV = cdt$
or, $\int {dV = \int {cdt} } $ or, $V = ct$ ..... (ii)
From Eqs. (i) and (ii), we get
${{dp} \over {dt}} = {{ - kc} \over {{c^2}{t^2}}}$
or, ${{dp} \over {dt}} = - {\left( {{k \over c}} \right)^{{t^{ - 2}}}}$
$ \Rightarrow dp = - {k \over c}.\,{t^{ - 2}}dt$
Integrating both sides, we get
$\int {dp = - {k \over c}\int {{t^{ - 2}}dt} } $
$p = - {k \over c}.\,\left( {{{{t^{ - 2 + 1}}} \over { - 2 + 1}}} \right)$
$ = - {k \over c}.\,{{ - 1} \over t} = {k \over {ct}}$
or, $p \propto {1 \over t}$
Hence, p versus ${1 \over t}$ graph is a straight line, which is correctly represented in option (b).
