Functions, 2ADV EQ-Bank 9

It is known that the volume of a hailstone \((V)\) is directly proportional to the cube of its radius \((r)\).

A hailstone with a radius of 1.25 cm has a volume of 8.2 cm\(^3\).

Find the equation relating \(V\) and \(r\). (2 marks)

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What is the expected radius of a hailstone with a volume of 51.1 cm\(^3\) ? Give your answer to 1 decimal place. (1 mark)

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a. \(V=4.1984 \times r^{3}\)

b. \(2.3\ \text{cm (2 d.p.)}\)

Show Worked Solution

a. \(V \propto r^3 \ \Rightarrow \ V=kr^3\)

\(\text{Find} \ k \ \text{given} \ \ V=8.2 \ \ \text{when}\ \ r=1.25:\)

\(8.2\)	\(=k \times 1.25^3\)
\(k\)	\(=\dfrac{8.2}{1.25^3}=4.1984\)

\(\therefore V=4.1984 \times r^{3}\)

b. \(\text{Find \(r\) when \(\ V=51.1\):}\)

\(51.1\)	\(=4.1984 \times r^3\)
\(r\)	\(=\sqrt[3]{\dfrac{51.1}{4.1984}}=2.3\ \text{cm (1 d.p.)}\)