 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2021/ib/paper-2/5a/a/solution

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

z (t)

is the water depth and

A (z) = π r (z)^{2}

is the horizontal cross-sectional area, Torricelli's law gives the volume flux through the hole as

q_{0} = π r_{0}^{2} 2 g z .

Conservation of volume gives

A (z) \overset{z}{˙} = - π r_{0}^{2} 2 g z .

For a prescribed constant fall rate

\overset{z}{˙} = - α

α > 0

, this requires

A (z) = \frac{π r _{0}^{2} 2 g}{α} z .

Consequently

r (z) = r_{0} (\frac{2 g}{α})^{1/2} z^{1/4},

so the container radius must be proportional to the fourth root of height.

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