 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ib/paper-3/18d/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Let

u (t)

be the signed air speed along the straw towards the bubble. Conservation of mass in a pipe for the incompressible plug flow gives

π ε^{2} u = \frac{d}{d t} (\frac{4 π a ^{3}}{3}), u = \frac{4 a ^{2} a ˙}{ε ^{2}} .

The unsteady Bernoulli equation along the straw, whose open end is at atmospheric pressure, gives

p_{in} - p_{atm} = - ρ L \overset{u}{˙} = \frac{2 γ}{a}

by the Young–Laplace equation. Substitution yields

a^{3} \overset{a}{¨} + 2 a^{2} \overset{a}{˙}^{2} = - \frac{γ ε ^{2}}{2 ρ L} = - C .

w = a^{2} \overset{a}{˙}

. Since

a \overset{w}{˙} = - C

, the chain rule gives

w d w / d a = - C a

\overset{a}{˙} (0) = 0

implies

a^{4} \overset{a}{˙}^{2} = C (a_{0}^{2} - a^{2}), a^{2} \overset{a}{˙} = - C (a_{0}^{2} - a^{2}) .

a = a_{0} sin θ

and integrating gives the implicit solution

t = \frac{a _{0}^{2}}{2 C} [\frac{π}{2} - θ + sin θ cos θ], θ = sin^{- 1} (a / a_{0}) .

At disappearance

a = 0

and

θ = 0

, so

t_{*} = \frac{π a _{0}^{2}}{2 ε} (\frac{ρ L}{2 γ})^{1/2} .

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