 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2023/paper-1/12/a/ii/solution

At each height, the removed circular sector has the same angle and

4/5

of the original radius. Its area and hence its volume are therefore multiplied by

(4/5)^{2} = 16/25

. Equating the removed volume to that of the reformed solid sphere gives

\frac{4}{3} π a^{3} = \frac{16}{25} V = \frac{2 π R ^{2} H}{25} (3 - e^{- 2}) .

Consequently

a = [\frac{3 R ^{2} H}{50} (3 - e^{- 2})]^{1/3} .

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