Classical Collapse

$Since\quad the\quad process\quad can\quad be\quad regarded\quad as\quad quasi-static,\quad we\quad can\quad reverse\quad time\quad i.e.\quad the\quad time\quad taken\quad by\quad the\quad bubble\quad to\quad collapse\quad is\quad the\quad same\quad as\quad the\quad time\quad taken\quad to\quad blow\quad a\quad bubble\quad of\quad the\quad same\quad dimension.\\ \\ Imagine\quad a\quad piston\quad pushing\quad air\quad with\quad constant\quad velocity\quad v\quad in\quad the\quad capillary\quad tube\quad of\quad cross-section\quad area\quad A\quad to\quad blow\quad a\quad bubble\quad of\quad radius\quad r\quad .\\ Consider\quad an\quad infinitesimally\quad small\quad cylindrical\quad element\quad of\quad air\quad in\quad the\quad capillary\quad tube.\\ The\quad length\quad of\quad this\quad element\quad is\quad vdt\quad and\quad its\quad mass\quad is\quad \rho Avdt\quad where\quad \rho \quad is\quad the\quad density\quad of\quad air.\\ Assuming\quad the\quad collision\quad between\quad the\quad air\quad molecules\quad and\quad the\quad surface\quad of\quad soap\quad bubble\quad to\quad be\quad inelastic,\\ the\quad force\quad exerted\quad on\quad the\quad surface\quad of\quad the\quad bubble\quad is\quad { F }_{ bubble }=\frac { dP }{ dt } =\frac { (\rho Avdt)v }{ dt } =\rho A{ v }^{ 2 }\\ The\quad total\quad distance\quad the\quad piston\quad travels\quad is\quad vT\quad \\ Using\quad Work\quad Energy\quad Theorem\quad and\quad excluding\quad the\quad piston\quad in\quad our\quad system,\quad we\quad get\\ { W }_{ ext }=\rho A{ v }^{ 2 }(vT)=\rho A{ v }^{ 3 }T\\ \Delta U=4\pi { r }^{ 2 }(2\sigma )=8\sigma \pi { r }^{ 2 }\quad where\quad \sigma \quad is\quad the\quad surface\quad tension\quad of\quad soap\quad bubble\\ \Delta T=\Delta \frac { 1 }{ 2 } m{ v }^{ 2 }=-\frac { 1 }{ 2 } (\rho AvT){ v }^{ 2 }=-\frac { \rho AT{ v }^{ 3 } }{ 2 } \quad as\quad the\quad final\quad kinetic\quad energy\quad in\quad the\quad molecules\quad is\quad 0\quad as\quad they\quad have\quad undergone\quad inelastic\quad collision.\\ Therefore\quad { W }_{ ext }\quad =\quad \Delta U\quad +\quad \Delta T\\ \rho A{ v }^{ 3 }T\quad =\quad 8\sigma \pi { r }^{ 2 }\quad -\quad \frac { \rho AT{ v }^{ 3 } }{ 2 } \\ Conserving\quad volume,\quad AvT=\frac { 4 }{ 3 } \pi { r }^{ 3 }\\ Combining\quad both\quad the\quad above\quad equations,\quad we\quad get\quad v=\sqrt { \frac { 4\sigma }{ \rho r } } \\ Substituting\quad this\quad in\quad Volume\quad Conservation\quad gives\quad T=\frac { 4\pi }{ 3A } \sqrt { \frac { \rho }{ 4\sigma } } { r }^{ 7/2 }=Z{ r }^{ 7/2 }\\ Hence\quad the\quad required\quad time\quad is\quad \bar { T } =Z(k{ r) }^{ 7/2 }=T{ k }^{ 7/2 }\\ Therefore,\boxed { n=\frac { 7 }{ 2 } }$

Relevant wiki: Bernoulli's Principle (Fluids)

Use Bernoulli theorem first of all. $Pin-Pout=$ $\frac{4*S}{R}$ in a soap bubble.Hence velocity $v^2$ = $\frac{C}{R}$ (Where $C$ is a constant).Eventually $v$ is proportional to $r^{-0.5}$ .Now change in volume or $dV$ = $4*\pi*r^2*dr$ .Also $dV=v*dt*A$ (where A is the surface area of capillary tube)....soon u will get that $r^2*dr$ proportional to $r^{-0.5}*dt$ ...now proceed to get that $r^{3.5}$ is proportional to $t$ .Hence the answer is $3.5$ and we r done :) . $LaTeX$