4MA0 Higher January 2013 Paper 4 Q21
21 When a photograph is taken, the exposure time, \(t\), is directly proportional to the square of the size, \(f\), of the opening in the camera lens.
\(t = 0.02\) when \(f = 8\)
(a) Find a formula for \(t\) in terms of \(f\). (3)
(b) Calculate the value of \(f\) when \(t = 0.0098\) (2)
| Scheme | Marks |
|---|---|
| \(t = kf^2\) | M1 |
\(0.02 = k \times 8^2\) or \(k = \dfrac{1}{3200}\) or \(k = 0.0003125\) or \(3.125 \times 10^{-4}\) | M1 |
| \(t = 0.0003125f^2\) or \(t = \dfrac{1}{3200}f^2\) | A1 |
| (3) |
Notes
M1: for \(t = kf^2\) but not for \(t = f^2\)
Also award for correct equation in \(t\), \(f^2\) and a constant or for \(t\) = some numerical value \(\times f^2\)
M1: for \(0.02 = k \times 8^2\) or for correct substitution into an equation which scores the first method mark (may be implied by correct evaluation of the constant)
A1: Award 3 marks if answer is \(t = kf^2\) but \(k\) is evaluated in part (b)
| Scheme | Marks |
|---|---|
| \(f^2 = \dfrac{0.0098}{0.0003125}\) or \(f^2 = \dfrac{0.0098}{0.02} \times 8^2\) | M1 |
| 5.6 oe | A1 |
| (2) |
Notes
M1: for substitution and rearrangement into form \(f^2 = \dfrac{0.0098}{k}\) with their value of \(k\) except for \(k = 1\) or \(f^2 = \dfrac{0.0098}{0.02} \times 8^2\)