Higher June 2025 Paper 2R Q26
26 The diagram shows a solid hemisphere, \(H\)
Diagram NOT accurately drawn
The radius of \(H\) is \(x\) cm
The volume of \(H\) is \(6174\pi\) cm3
A bowl is made by removing a solid hemisphere from \(H\) such that the uniform thickness of the bowl is 2 cm
Diagram NOT accurately drawn
Work out the total surface area of the bowl.
Give your answer in terms of \(\pi\)
(5)
| Scheme | Marks |
|---|---|
| eg \(\dfrac{1}{2} \times \dfrac{4}{3}\pi x^3 = 6174\pi\) | M1 |
| \((x =)\sqrt[3]{\dfrac{6174\pi \times 3}{2\pi}}\;\left(= \sqrt[3]{9261} = 21\right)\) | M1 |
| eg \(\pi([21])^2 - \pi([21] - 2)^2\;(= 441\pi - 361\pi = 80\pi)\) oe or \(2\pi([21])^2 + 2\pi([21] - 2)^2\;(= 882\pi + 722\pi = 1604\pi)\) oe | M1ft |
| \(\text{``}{1604\pi}\text{''} + \text{``}{80\pi}\text{''}\) | M1ft |
| Correct answer only scores full marks (unless from obviously incorrect working) Answer: \(1684\pi\) | A1 |
| (5) | |
| (5 marks) |
Notes
M1: oe
for forming a correct equation; allow use of any letter
M1: for a correct method to find the radius of the hemisphere
M1ft: for a method to find the area of the top of the bowl
or the total area of the two curved surfaces of the bowl
where [21] is what they believe to be radius of the hemisphere
M1ft: ft their [21]
for a complete method
A1: cao
SCB4 for \(2028\pi\) (use of 23 as the outer radius)
M2 for use of formula for total surface area of hemispherical shell in a complete method,
eg \(3\pi([21])^2 + \pi([21] - 2)^2\) oe
If not M2, allow M1 for this formula used with omission of \(\pi\)
eg \(3([21])^2 + ([21] - 2)^2\)