Higher June 2021 Paper 2 Q16
16 \(A\) is inversely proportional to the square of \(r\)
\(A = 5\) when \(r = 0.3\)
(a) Find a formula for \(A\) in terms of \(r\) (3)
(b) Find the value of \(A\) when \(r = 7.5A\) (3)
| Scheme | Marks |
|---|---|
| \(A = \dfrac{k}{r^2}\) | M1 |
| \(5 = \dfrac{k}{0.3^2}\) oe or \(k = 0.45\) oe | M1 |
Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: \(A = \dfrac{0.45}{r^2}\) | A1 |
| (3) |
Notes
M1: oe \(k\) can be any letter (must be a letter and not 1)
M1: implies first M1 if you see this stage
A1: oe with \(A\) as the subject eg \(A = \dfrac{9}{20r^2}\)
(allow \(A = \dfrac{k}{r^2}\) where \(k = 0.45\) oe)
(SC if M0 scored then award B2 for \(A \propto \dfrac{0.45}{r^2}\) oe)
| Scheme | Marks |
|---|---|
\([A =]\ \dfrac{\text{``}{0.45}\text{''}}{(7.5A)^2}\) oe or \(\dfrac{\text{``}{0.45}\text{''}}{56.25A^2}\) or \(\dfrac{9}{20(7.5A)^2}\) oe | M1 |
\(A^3 = \dfrac{\text{``}{0.45}\text{''}}{56.25}\) (\(A^3 = \dfrac{1}{125}\) or 0.008 oe) or \(125A^3 = 1\) oe | M1 |
| Working not required, so correct answer scores full marks (unless from obvious incorrect working) Answer: 0.2 | A1 |
| (3) | |
| (6 marks) |
Notes
M1: ft from (a) dep on M2 in (a)
(\([A =]\ \dfrac{\text{``}{0.45}\text{''}}{7.5A^2}\) is zero marks unless recovered later)
M1: ft their 0.45 dep on M2 in (a)
Must include \(A^3\)
A1: oe