QJEEM19Ph8ff2: JEE Main 2019 Physics - Class 11 JEEM19Ph8ff2 Solution: A soap bubble, blown by a mechanical pump at the mouth of a…

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).