 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-4/39a/c/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

p_{1} - p_{0} = ρ_{0} V^{2} - \frac{3}{2} ρ_{0} (\frac{2 V}{3})^{2} = \frac{1}{3} ρ_{0} V^{2} .

For

γ = 3

, energy conservation becomes

\frac{3 p _{0}}{2 ρ _{0}} + \frac{1}{2} V^{2} = \frac{p _{1}}{ρ _{0}} + \frac{2}{9} V^{2} .

Eliminating

ρ_{0} V^{2}

between these equations yields

p_{1} = 4 p_{0}

. Hence

\frac{p _{1}}{p _{0}} = 4 .

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