Uncertainties, graphs & data-based questions – 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.

3Hard6 marks

Students measure the period T of a simple pendulum for different lengths l. The graph shows T² against l. The line of best fit passes through (0.10, 0.40) and (0.90, 3.62).

(a) Explain why plotting T² against l should give a straight line through the origin. [1]

(b) Determine the gradient of the line of best fit. [2]

(d) The steepest and shallowest acceptable lines have gradients 4.11 and 3.95 s² m⁻¹. Determine the absolute uncertainty in g. [1]

(a) T = 2π√(l/g) so T² = (4π²/g)·l: T² is directly proportional to l (form y = mx, no intercept) (OWTTE)(1)
(b) Uses two points on the line (not data points), well separated: (3.62 − 0.40)/(0.90 − 0.10)(1)
(b) Gradient = 4.03 s² m⁻¹ (accept 4.02 or 4.0)(1)
(c) g = 4π²/gradient(1)
(c) g = 9.81 ≈ 9.8 m s⁻² (allow 9.8–9.9; allow ECF from (b))(1)
(d) Half-range of gradients = 0.08 (2.0%), so Δg ≈ 2% of 9.8 = ± 0.2 m s⁻² (accept (4π²/3.95 − 4π²/4.11)/2 = 0.19 m s⁻²; accept 0.19–0.2; allow ECF from (c))(1)
Answer: (a) $T^2=(4\pi^2/g)\,l$, so $T^2\propto l$; (b) 4.03 s² m⁻¹ (4.0); (c) 9.8 m s⁻²; (d) ± 0.2 m s⁻²
Examiner tip: Take the gradient from points on the best-fit line, not from the plotted data points.

4HardHL only5 marks

The activity A of a sample of a radioisotope is measured over time t. The graph shows ln(A / Bq) against t / h. The line of best fit passes through (0, 9.80) and (6.0, 8.41).

(a) Show that the data are consistent with A = A₀e^−λt by deriving a linear relationship between ln A and t. [1]

(b) Determine λ. [2]

(d) Determine the initial activity. [1]

(a) Taking natural logs: ln A = ln A₀ − λt, a straight line with gradient −λ and intercept ln A₀(1)
(b) Gradient = (8.41 − 9.80)/6.0 using points on the line(1)
(b) λ = 0.232 h⁻¹ (magnitude; accept 0.23 h⁻¹ or 6.4 × 10⁻⁵ s⁻¹; unit required)(1)
(c) T½ = ln2/λ = 2.99 h ≈ 3.0 h (allow ECF from (b); accept 1.08 × 10⁴ s)(1)
(d) A₀ = e^9.80 = 1.80 × 10⁴ Bq (allow 1.8 × 10⁴)(1)
Answer: (b) 0.232 h⁻¹; (c) 3.0 h; (d) 1.8 × 10⁴ Bq
Examiner tip: The gradient is −λ, so λ is positive; the intercept is ln A₀, not A₀ itself.

5Standard5 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.

6Hard5 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.

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