Higher June 2025 Paper 1R Q18
18 \(F\) is inversely proportional to the cube of \(r\)
\(F = 6\) when \(r = 2\)
(a) Find a formula for \(F\) in terms of \(r\) (3)
(b) Find the value of \(r\) when \(F = 3072\) (2)
| Scheme | Marks |
|---|---|
| \(F = \dfrac{k}{r^3}\) or \(Fr^3 = k\) or \(kF = \dfrac{1}{r^3}\) | M1 |
\(6 = \dfrac{k}{2^3}\) oe or \(k = 48\) or \(6k = \dfrac{1}{2^3}\) oe or \(k = \dfrac{1}{48}\) | M1 |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(F = \dfrac{48}{r^3}\) | A1 |
| (3) |
Notes
M1: oe \(k\) can be any letter (must be a letter and not 1)
M1: For substitution of \(F\) and \(r\) into a correct formula, implies the first M1 if you see this stage
Condone use of \(\propto\) for method marks
A1: oe with \(F\) the subject eg \(F = 48 \times \dfrac{1}{r^3}\) or
\(F = 48 \times r^{-3}\)
Award 3 marks if answer is \(F = \dfrac{k}{r^3}\) and
\(k = 48\) clearly given in the body of the script
M2A0 for \(Fr^3 = 48\) or \(r = \sqrt[3]{\dfrac{48}{F}}\) or \(r^3 = \dfrac{48}{F}\)
| Scheme | Marks |
|---|---|
| \((r^3 =)\dfrac{\text{``}{48}\text{''}}{3072}\) oe eg \(\dfrac{1}{64}\) or (0.01(5625)) rounded or truncated | M1ft |
Correct answer scores full marks (unless from obvious incorrect working) Answer: \(\dfrac{1}{4}\) | A1 |
| (2) | |
| (5 marks) |
Notes
M1ft: allow use of their “48” as long as M2 gained in (a)
A1: oe