Higher January 2022 Paper 2R Q21
21 The diagram shows a solid made from a cylinder and a hemisphere.
The cylinder and the hemisphere are both made from the same metal.
Diagram NOT accurately drawn
The plane face of the hemisphere coincides with the upper plane face of the cylinder.
The radius of the cylinder and the radius of the hemisphere are both \(x\) cm.
The height of the cylinder is \(3x\) cm.
The total surface area of the solid is \(81\pi\) cm2
The mass of the solid is 840 grams.
The following table gives the density of each of four metals.
| Metal | Density (g/cm3) |
|---|---|
| Aluminium | 2.7 |
| Nickel | 8.9 |
| Gold | 19.3 |
| Silver | 10.5 |
The metal used to make the solid is one of the metals in the table.
Determine the metal used to make the solid.
Show your working clearly.
(6)
| Scheme | Marks |
|---|---|
\(\pi x^2 + 2\pi x \times 3x + \dfrac{1}{2} \times 4\pi x^2 = 81\pi\) oe or \(9x^2 = 81\) oe or \(2\pi x \times 3x + \dfrac{1}{2} \times 4\pi x^2 = 81\pi\) oe or \(8x^2 = 81\) | M1 |
| \((x =)\ \sqrt{\dfrac{81}{9}}\ (= 3)\) | M1 |
\(\pi \times \text{‘}{3}\text{’}^2 \times 3 \times \text{‘}{3}\text{’} + \dfrac{1}{2} \times \dfrac{4}{3}\pi \text{‘}{3}\text{’}^3\) oe (\(= 81\pi + 18\pi = 99\pi = 311.(017\ldots)\)) | M1 |
| \(99\pi\) or 311.(017…) | A1 |
| \(\dfrac{840}{\text{‘}{311}\text{’}}\ (= 2.7\ldots)\) oe | M1 |
| aluminium | A1 |
| (6) | |
| (6 marks) |
Notes
M1: for setting up an equation (in a single variable ie \(x\) or \(r\)) for the total surface area of the shape or for the curved surface area.
M1: solving their equation in the form \(kx^2\pi = 81\pi\) (where \(k\) follows correctly from their surface area) to find \(x\)
M1: (indep) for substituting their value of \(x\) to find the volume of the shape.
M1: (dep on the 3rd M) for using the formula for density
A1: for aluminium and correct working leading to 2.7