A student is investigating a rotary variable resistor, as shown in Fig. 2.1. [Figure 2.1] The variable resistor is connected to a battery of electromotive force (e.m.f.) E and negligible internal resistance, as shown in Fig. 2.2. [Figure 2.2] The student uses a protractor to measure the angle θ through which the spindle of the variable resistor is rotated and records the current I. The experiment is repeated for different angles. It is suggested that I and θ are related by the equation E = IKθ where K is a constant.

Mark scheme

Marking points from the official mark scheme (May/June 2019). The mark or mark type is shown in brackets.

1. [1] gradient = K / E
2. [1] 1/I / A⁻¹ values: 180 or 175, 210 or 213, 250, 290 or 286, 320 or 323, 370
3. [1] uncertainties in 1/I from ±3 or ±4 to ±10–15
4. [1] Six points plotted correctly. Must be accurate to the nearest half small square. Diameter of points must be less than half a small square.
5. [1] Error bars in 1/I plotted correctly. All error bars to be plotted. Length of bar must be accurate to less than half a small square and symmetrical.
6. [1] Line of best fit drawn. Do not allow line from top point to bottom point. If points are plotted correctly then lower end of line should pass between (122, 230) and (126, 230) and upper end of line should pass between (186, 350) and (190, 350).
7. [1] Worst acceptable line drawn (steepest or shallowest possible line that passes through all the error bars). All error bars must be plotted.
8. [1] Gradient determined with clear substitution of points from the line into Δy/Δx. Distance between points must be at least half the length of the drawn line.
9. [1] uncertainty = gradient of line of best fit – gradient of worst acceptable line or uncertainty = 1/2 (steepest worst line gradient – shallowest worst line gradient)
10. [1] 9.4 ± 0.2 (V)
11. [1] K determined from gradient and given to 2 or 3 significant figures. K = E x gradient = 9.4 × (c)(iii).
12. [1] K determined from gradient with correct unit (Ω/°).
13. [1] % uncertainty in K = ( (Δgradient / gradient) + (ΔE / E) ) x 100
14. [1] θ calculated. Correct substitution of numbers required. θ = E / (IK) = 9.4 / (0.01 × (e)(i)) OR θ = 1 / (I × gradient) = 1 / (0.01 × (c)(iii))
15. [1] Absolute uncertainty in θ. Correct substitution of numbers required. Use of ΔI not required but allow if included by the candidate. Using E and K: uncertainty in θ = ( (ΔE / E) + (ΔK / K) + (ΔI / I) ) × θ OR uncertainty in θ = ( (0.2 / 9.4) + (e)(ii) / 100 ) × θ OR max θ = max E / (0.01 × min K) OR min θ = min E / (0.01 × max K). Using gradient: uncertainty in θ = ( (Δgradient / gradient) + (ΔI / I) ) × θ OR max θ = 1 / (0.01 × min gradient) OR min θ = 1 / (0.01 × max gradient)