Higher January 2020 Paper 1R Q20
20 The diagram shows a frustum of a cone and a sphere.
The frustum is made by removing a small cone from a large cone.
The cones are similar.
Diagram NOT accurately drawn
The height of the small cone is \(h\) cm.
The height of the large cone is \(2h\) cm.
The radius of the base of the large cone is \(r\) cm.
The radius of the sphere is \(r\) cm.
Given that the volume of the frustum is equal to the volume of the sphere,
find an expression for \(r\) in terms of \(h\).
Give your expression in its simplest form.
(5)
| Scheme | Marks |
|---|---|
| \(\dfrac{1}{3} \times \pi \times r^2 \times 2h \left(= \dfrac{2}{3}\pi r^2 h\right)\) OR \(\dfrac{1}{3} \times \pi \times (0.5r)^2 \times h \left(= \dfrac{1}{12}\pi r^2 h\right)\) | M1 |
| \(\text{“}\dfrac{2}{3}\pi r^2 h\text{”} - \text{“}\dfrac{1}{12}\pi r^2 h\text{”} \left(= \dfrac{7}{12}\pi r^2 h\right)\) | M1 |
| \(\text{“}\dfrac{2}{3}\pi r^2 h\text{”} - \text{“}\dfrac{1}{12}\pi r^2 h\text{”} = \dfrac{4\pi r^3}{3}\) | M1 |
| e.g. \(\dfrac{7}{12}\pi r^2 h = \dfrac{4\pi r^3}{3}\) | M1 |
| \(\dfrac{7}{16}h\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: for finding the volume of the small or large cone
M1: (dep) method to find the volume of the frustum
(condone missing brackets)
M1: equating volume of frustum and sphere (must be correct including brackets)
M1: for a correct simplified formula (1 term on each side)
A1: accept \(0.4375h\)