 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ii/paper-2/39c/b/solution

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

u_{r} = (p_{r} /2 μ) z (z - h)

and continuity gives

\int_{0}^{h} u_{r} d z = - V r /2

. Hence

p_{r} = 6 μ V r / h^{3}

and

p - p_{0} = 3 μ V (r^{2} - a^{2}) / h^{3}

. Balancing the upward pressure force with

W

yields

\dot{h} = - \frac{2 W h ^{3}}{3 π μ a ^{4}} .

Therefore $h^{- 2} = h_{0}^{- 2} + 4 W t / (3 π μ a^{4})$ and $h \sim (3 π μ a^{4} /4 W t)^{1/2}$ .

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