Higher November 2017 Unit 1 Q10
10. A cylinder just fits inside a hollow cube with sides of length \(m\) cm.
Diagram not drawn to scale
The radius of the cylinder is \(\dfrac{m}{2}\) cm.
The height of the cylinder is \(m\) cm.
The ratio of the volume of the cube to the volume of the cylinder is given by
volume of cube : volume of cylinder
= \(k : \pi\),
where \(k\) is a number.
Find the value of \(k\).
You must show all your working. [4]
| Scheme | Marks | Comments |
|---|---|---|
| (Volume of cube =) \(m^3\) OR \(m \times m \times m\) OR \(m^2 \times m\) | B1 | For sight of \(m^3\) or equivalent. |
| (Volume of cylinder =) \(\dfrac{\pi m^3}{4}\) OR \(\dfrac{\pi \times m \times m \times m}{4}\) OR \(\dfrac{\pi \times m^2 \times m}{4}\) | B2 | For sight of \(\pi m^3/4\) or equivalent. B1 for \(\pi \times \left(\dfrac{m}{2}\right)^2 \times m\). Also allow this B1 if brackets are missing. \(m^3 : \dfrac{\pi m^3}{4}\) OR \(4m^3 : \pi m^3\) OR \(1 : \dfrac{\pi}{4}\) all imply B1B2. |
| \(k = 4\) | B1 | Allow B1 if left as 4 : π . F.T. only for \(\pi m^3 / 2\) (giving \(k\) = 2 or 2 : π) Note : If a value is used for m then mark as above and penalise −1 from total mark gained. |