Measurement, errors and uncertainties – IB Physics exam questions with mark scheme

1Core3 marks

A rectangle has measured sides of (12.0 ± 0.2) cm and (8.0 ± 0.2) cm.

Calculate the area of the rectangle with its absolute uncertainty. [3]

Percentage uncertainties: 0.2/12.0 = 1.7% and 0.2/8.0 = 2.5%(1)
Percentage uncertainties added for a product: 4.2%(1)
Area = (96 ± 4) cm²(1)
Answer: (96 ± 4) cm²
Examiner tip: Add percentage uncertainties when multiplying, then convert back to an absolute uncertainty.

2Standard4 marks

A trolley of mass (0.250 ± 0.005) kg moves at a speed of (4.0 ± 0.2) m s⁻¹.

Calculate its kinetic energy and the absolute uncertainty, giving the final answer in the form E ± ΔE. [4]

Kinetic energy = ½mv² = 2.0 J(1)
Fractional uncertainties: m is 2% and v is 5%(1)
v is squared so its percentage uncertainty is doubled (10%); total = 2% + 10% = 12%(1)
ΔE = 0.12 × 2.0 = 0.24 J, so E = (2.0 ± 0.2) J (uncertainty to 1 s.f.)(1)
Answer: (2.0 ± 0.2) J
Examiner tip: The uncertainty in a squared quantity is twice its percentage uncertainty; quote ΔE to 1 s.f. and match decimal places.

3Standard5 marks

In an experiment to find the resistance of a resistor, a student records the current I through it for several values of potential difference V. After plotting the results, the student finds that the ammeter reads 0.05 A when no current flows. The recorded readings were not corrected for this.

(a) Identify the type of error and state its effect on a graph of I against V. [2]

(b) State one way to reduce the effect of random errors in the readings. [1]

(a) Systematic (zero / offset) error (reject “random”)(1)
(a) Every current reading is 0.05 A too high, so the line is shifted up by 0.05 A: intercept of +0.05 A on the I axis / line does not pass through the origin (accept line parallel to the true line; stating the gradient is unchanged is not required but must not be contradicted)(1)
(b) Repeat each measurement and calculate a mean (also accept: take more data points and use a line of best fit)(1)
(c) Draw the steepest and the shallowest straight lines that still pass through all the error bars (accept max and min gradient lines through the error bars)(1)
(c) Uncertainty in gradient = half the difference between their gradients, (max − min)/2 (accept the difference between the best-fit gradient and the max or min gradient)(1)
Answer: (a) systematic (zero) error: line shifted up by 0.05 A (intercept +0.05 A), gradient unchanged; (b) repeat readings and average; (c) draw max and min gradient lines through all error bars; uncertainty = (max − min)/2
Examiner tip: A zero error is systematic, not random; it moves the line but does not change its gradient.

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4Hard5 marks

The resistivity ρ of a wire is found from ρ = RA/L. Measurements: resistance R = (2.00 ± 0.05) Ω; length L = (1.000 ± 0.002) m; diameter d = (0.52 ± 0.01) mm.

(a) Calculate the resistivity of the wire. [2]

(b) Determine the percentage uncertainty in ρ. [2]

(a) Cross-sectional area A = πd²/4 = π(0.52 × 10⁻³)²/4 = 2.12 × 10⁻⁷ m²(1)
(a) ρ = RA/L = 4.25 × 10⁻⁷ Ω m(1)
(b) Percentage uncertainties: R 2.5%, d 1.9% (doubled to 3.8% as A ∝ d²), L 0.2%(1)
(b) Total = 6.5% (≈ 6.5%)(1)
(c) ρ = (4.2 ± 0.3) × 10⁻⁷ Ω m (uncertainty to 1 s.f., value matched)(1)
Answer: ρ = (4.2 ± 0.3) × 10⁻⁷ Ω m
Examiner tip: Remember d appears squared, so double its percentage uncertainty, and quote the uncertainty to one significant figure.

5Core4 marks

A student uses a light gate to measure the time t for a ball to fall a fixed height. Five repeat readings are taken:

t / s	0.452	0.448	0.461	0.439	0.455

(a) Calculate the mean value of t. [1]

(b) Estimate the absolute uncertainty in the mean value of t. [1]

(d) The light gate records times to 0.001 s. Explain why the uncertainty in (b) is larger than the resolution of the light gate. [1]

(a) mean $t$ = 0.451 s (allow 0.4510 s)(1)
(b) half the range: $\Delta t = \dfrac{0.461-0.439}{2} = 0.011$ s (allow 0.01 s; do not accept 0.001 s, the resolution)(1)
(c) $t = (0.45 \pm 0.01)$ s (allow (0.451 ± 0.011) s; the value must be given to the same decimal place as the uncertainty; ECF from (a) and (b))(1)
(d) the spread of repeat readings is caused by random errors (e.g. variation in release / position of the ball) that are larger than the instrument resolution (accept 'the readings vary by more than 0.001 s, so random variation dominates')(1)
Answer: (a) 0.451 s (b) ±0.011 s (c) t = (0.45 ± 0.01) s (d) random variation between repeats is bigger than the resolution.
Examiner tip: Many students quote the light-gate resolution (±0.001 s) as the uncertainty: when repeats spread out, use half the range instead.

6Core3 marks

Newton's law of gravitation is $F = \dfrac{Gm_1m_2}{r^2}$.

(a) Show that the SI base units of the gravitational constant G are kg⁻¹ m³ s⁻². [2]

(b) The diameter of the Earth is about 1.3 × 10⁷ m and the diameter of an atom is about 1 × 10⁻¹⁰ m. Determine the order of magnitude of the ratio $\dfrac{\text{diameter of Earth}}{\text{diameter of atom}}$. [1]

(a) rearranges to $G = \dfrac{Fr^2}{m_1m_2}$ and writes N as kg m s⁻²(1)
(a) substitutes units: $\dfrac{\text{kg m s}^{-2}\,\text{m}^2}{\text{kg}^2}$ = kg⁻¹ m³ s⁻² (answer is given, so the working must be shown; leaving the newton unconverted scores only the first mark at most)(1)
(b) 10¹⁷ (ratio = 1.3 × 10¹⁷; do not accept 1.3 × 10¹⁷ on its own, an order of magnitude is a power of ten)(1)
Answer: (a) G = Fr²/(m₁m₂) → kg m s⁻² × m² / kg² = kg⁻¹ m³ s⁻² (b) 10¹⁷
Examiner tip: In 'show that' unit questions, expand the newton into kg m s⁻² explicitly: just quoting the given answer gets nothing.

7Standard5 marks

A student determines the acceleration of free fall g using a simple pendulum. The length is measured as L = (0.652 ± 0.002) m. The time for 20 complete oscillations is (32.4 ± 0.4) s.

(a) Calculate the period T and its absolute uncertainty. [1]

(b) Explain why the student timed 20 oscillations rather than one. [1]

(a) $T = 1.62$ s and $\Delta T = 0.02$ s(1)
(b) the (absolute) uncertainty in the timing (e.g. reaction time) is spread over 20 oscillations, so the percentage / fractional uncertainty in T is reduced (by a factor of 20) (do not accept 'more accurate' alone)(1)
(c) $g = \dfrac{4\pi^2 \times 0.652}{1.62^2} = 9.81$ m s⁻²(1)
(c) fractional uncertainty: $\dfrac{0.002}{0.652} + 2 \times \dfrac{0.02}{1.62}$ = 0.0031 + 0.0247 = 0.028 / 2.8 % (the factor 2 on T is required for this mark)(1)
(c) $\Delta g$ = 0.028 × 9.81 ≈ 0.3 m s⁻², so $g = (9.8 \pm 0.3)$ m s⁻² (allow (9.81 ± 0.27) m s⁻²; ECF from incorrect fractional uncertainty)(1)
Answer: (a) T = (1.62 ± 0.02) s (c) g = (9.8 ± 0.3) m s⁻²
Examiner tip: The squared T doubles its percentage uncertainty: students who simply add 0.3 % and 1.2 % lose the method mark.

8Hard6 marks

A student investigates how the extension x of a spring depends on the load F. Each extension has an absolute uncertainty of ±0.3 cm; the uncertainty in F is negligible. The graph shows the data with error bars and the line of best fit, which passes through (0, 0.6) and (5.0, 10.5).

(a) Calculate the percentage uncertainty in the extension when F = 1.0 N. [1]

(b) The student claims that x is directly proportional to F. Explain, with reference to the error bars, whether the data support this claim. [2]

(d) Determine the spring constant k of the spring, in N m⁻¹. [2]

(a) $\dfrac{0.3}{2.6} \times 100$ = 12 % (allow 11.5 %; allow 11–13 % from own reading of x ≈ 2.6 cm)(1)
(b) for direct proportionality the line must be straight and pass through the origin / no straight line through the origin passes through all the error bars (e.g. a line through the origin and the F = 1.0 N bar is too steep to pass through the F = 5.0 N bar)(1)
(b) the best-fit line is straight (within error bars) but has an intercept of about 0.6 cm on the extension (vertical) axis, which is larger than the uncertainty of 0.3 cm, so the data do not support direct proportionality (the conclusion needs the reason)(1)
(c) systematic error, e.g. the ruler zero not aligned with the bottom of the unloaded spring / the extension measured from the wrong starting length / a load (e.g. hanger) already on the spring not included in F / zero error (do not accept 'random error' or 'parallax' with no systematic direction)(1)
(d) gradient = $\dfrac{10.5-0.6}{5.0}$ = 1.98 cm N⁻¹ = 0.0198 m N⁻¹ (allow 1.9–2.0 cm N⁻¹ from own reading)(1)
(d) $k = \dfrac{1}{\text{gradient}}$ = 51 N m⁻¹ (allow 50–53 N m⁻¹; award 1 mark max for 0.51 (no cm→m conversion))(1)
Answer: (a) 12 % (b) No: straight but intercept ≈ 0.6 cm, larger than the error bars allow; no line through the origin fits all bars. (c) Systematic / zero error in measuring x. (d) k ≈ 51 N m⁻¹
Examiner tip: A straight line is not enough for proportionality: you must say whether a line through the origin fits within the error bars.

9Stretch7 marks

A student determines the specific heat capacity c of water. An immersion heater of power P = (48.0 ± 0.5) W heats a mass m = (0.250 ± 0.002) kg of water for a time t = (120 ± 1) s. A thermometer reads the initial temperature as (21.5 ± 0.5) °C and the final temperature as (26.5 ± 0.5) °C.

(a) Calculate c. [1]

(b) Determine the absolute uncertainty in c. [3]

(c) The accepted value is 4180 J kg⁻¹ K⁻¹. Comment on whether the experimental result is consistent with this value. [1]

(d) Suggest and explain one change to the procedure that would most reduce the uncertainty in c. [2]

(a) $c = \dfrac{Pt}{m\Delta T} = \dfrac{48.0 \times 120}{0.250 \times 5.0}$ = 4600 J kg⁻¹ K⁻¹ (allow 4608)(1)
(b) uncertainty in temperature change: $\Delta(\Delta T)$ = 0.5 + 0.5 = 1.0 K, giving 20 % (uncertainties add when quantities are subtracted)(1)
(b) total percentage uncertainty = 1.0 % + 0.8 % + 0.8 % + 20 % ≈ 23 % (allow 22.7 %; ECF from wrong ΔT uncertainty)(1)
(b) $\Delta c$ ≈ 0.23 × 4600 ≈ 1000 J kg⁻¹ K⁻¹, so $c = (4.6 \pm 1.0) \times 10^3$ J kg⁻¹ K⁻¹ (allow ±1.0 × 10³ or ±1045; ECF)(1)
(c) consistent: 4180 lies within the range 3600–5600 J kg⁻¹ K⁻¹ / the difference from 4180 is smaller than the uncertainty (ECF from (b); a judgement without reference to the range scores 0)(1)
(d) heat for longer / use a larger temperature rise (e.g. about 20 K) (accept: use a thermometer of higher resolution, e.g. ±0.1 °C / a temperature probe)(1)
(d) the temperature-change uncertainty is the dominant (largest) one, and a larger ΔT (or smaller ±) makes its percentage uncertainty much smaller (do not accept 'repeat the experiment' or 'more accurate thermometer' without a reason linked to ΔT)(1)
Answer: (a) c ≈ 4600 J kg⁻¹ K⁻¹ (b) ±1.0 × 10³ J kg⁻¹ K⁻¹ (23 %) (c) consistent, 4180 lies within 3600–5600 (d) larger temperature rise or finer thermometer, as ΔT dominates.
Examiner tip: When two readings are subtracted, add their absolute uncertainties first: a 5 K rise read to ±0.5 °C at each end carries 20 %, which swamps everything else.

10Hard6 marks

A student determines the density of a small irregular stone. The mass is measured with a top-pan balance that reads to 0.1 g. The volume is found by lowering the stone into a 100 cm³ measuring cylinder, graduated every 1 cm³, partly filled with water. The stone has a volume of about 8 cm³.

Discuss the errors and uncertainties in this experiment and how the student could reduce them to obtain a more accurate and precise value for the density. [6]

percentage uncertainty in mass is small (e.g. 0.1 g in about 20 g ≈ 0.5 %), so mass is not the main concern(1)
volume found from the difference of two cylinder readings: uncertainty about ±1 cm³ (±0.5 cm³ each) in about 8 cm³, i.e. over 10 %; this dominates(1)
percentage uncertainty in density = % in mass + % in volume(1)
use a narrower measuring cylinder with finer graduations (e.g. 10 or 25 cm³ read to 0.1–0.2 cm³) or a displacement (eureka) can with a small measuring cylinder; or use a larger stone sample(1)
read the bottom of the meniscus at eye level to avoid parallax (systematic) error(1)
zero / tare the balance before use and check zero error (systematic error is not reduced by repeating)(1)
dry the stone before weighing (water would add mass); remove air bubbles clinging to the stone in the water (would increase the measured volume)(1)
repeat readings and average to reduce random error; random errors affect precision, systematic errors affect accuracy(1)

Level	Marks	What it looks like
Level 3	5–6	Identifies the volume measurement as the dominant uncertainty with a quantitative or clear comparison to the mass, and gives several well-explained, specific improvements covering both random and systematic errors. Reasoning is logical and linked to accuracy versus precision.
Level 2	3–4	Describes at least two relevant errors or uncertainties and gives matching improvements, with some explanation. Comparison of the sizes of uncertainties may be missing or weak.
Level 1	1–2	Makes simple relevant points, such as 'repeat and average' or 'read the meniscus at eye level', with little explanation or linkage.
Level 0	0	No relevant content.

Answer: Volume uncertainty (over 10 %) dominates mass (about 0.5 %); use finer volume measurement / larger sample, read meniscus at eye level, tare the balance, dry the stone, remove bubbles, repeat and average.
Examiner tip: Top answers identify which measurement has the largest percentage uncertainty and target improvements there, rather than listing generic 'repeat it' advice.

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