Higher November 2018 Paper 2 Q20
20 Here is a frustum of a cone.
| Volume of sphere \(= \dfrac{4}{3}\pi r^3\) |
| Volume of cone \(= \dfrac{1}{3}\pi r^2 h\) |
The diagram shows that the frustum is made by removing a cone with height 3.2 cm from a solid cone with height 6.4 cm and base diameter 7.2 cm.
The frustum is joined to a solid hemisphere of diameter 7.2 cm to form the solid S shown below.
The density of the frustum is 2.4 g/cm\(^3\)
The density of the hemisphere is 4.8 g/cm\(^3\)
Calculate the average density of solid S. (5)
| Answer | Mark | Mark scheme |
|---|---|---|
| 3.75 | P1 | works to find vol of frustum eg \(1/3\pi(3.6)^2 \times 6.4 - 1/3\pi(1.8)^2 \times 3.2\) or \(86.858.. - 10.857\ldots\) \((= 24.192\pi\) or \(76.00..)\) |
| P1 | works to find vol of hemisphere eg \(\dfrac{1}{2} \times \dfrac{4}{3}\pi \times 3.6^3\) \((= 31.104\pi\) or \(97.7....)\) | |
| P1 | mass of frustum as [vol]\(\times\)density eg \(\text{``}76.00\text{''} \times 2.4\ (= 182.4..)\) or mass of hemisphere as [vol]\(\times\)density eg \(\text{``}97.7....\text{''} \times 4.8\ (= 469.037...)\) | |
| P1 | mean density as total mass \(\div\) total volume eg \((\text{``}182.4..\text{''} + \text{``}469.037\text{''}) \div (\text{``}76\ldots\text{''} + \text{``}97.7..\text{''})\) or \(\text{``}651.4..\text{''} \div \text{``}173.7....\text{''}\) | |
| A1 | answer in the range 3.7 to 3.8 |
Additional guidance
781.7… by use of diameter does not get the mark
[vol] is their volume which could be ft using the radius, using the diameter, or could be another value as long as it is stated as being the volume, or clearly intended from working.
All figures must come from correct method shown.