Graphs and data analysis – IB Physics exam questions with mark scheme

1Hard6 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.

2HardHL 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.

3Core3 marks

A student measures the intensity $I$ of light at different distances $d$ from a small lamp. The student predicts that the lamp emits light equally in all directions, so that

$$I = \frac{P}{4\pi d^{2}}$$

where $P$ is the power emitted as light.

(a) State the quantity that should be plotted on the horizontal axis, with $I$ on the vertical axis, to give a straight line through the origin. [1]

(b) State an expression for the gradient of this line. [1]

(a) $\dfrac{1}{d^{2}}$ (accept $d^{-2}$; do not accept $d^{2}$ or $\dfrac{1}{d}$)(1)
(b) $\dfrac{P}{4\pi}$ (accept gradient $= P/(4\pi)$ in words)(1)
(c) $P = 4\pi \times 0.27 = 3.4\ \mathrm{W}$ (accept 3.39 W; ECF from (b))(1)
Answer: (a) $1/d^{2}$ (b) $P/4\pi$ (c) 3.4 W
Examiner tip: Match the graph to $y = mx$: whatever multiplies the plotted variable is the gradient, so here the gradient is $P/4\pi$, not $P$.

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4Core4 marks

A student hangs loads on a spring and measures the extension $x$ for each force $F$. The extension is measured with a metre rule and has an uncertainty of $\pm 0.003$ m. The graph shows the results with error bars and the student's line of best fit.

(a) Identify the anomalous data point and state how the student should treat it. [2]

(b) Suggest why the student has drawn error bars for $x$ but not for $F$. [1]

(a) the point at $F = 2.0$ N / $x = 0.094$ m (accept a clear description, e.g. "the fourth point")(1)
ignore it when drawing the line of best fit / repeat that measurement (do not accept "delete it from the table" without repeat or exclusion from the line; ignore "it is an outlier" alone)(1)
(b) the uncertainty in $F$ is negligible / very small compared with that in $x$ OR the masses/weights are accurately known (allow error bar in $F$ would be too small to show)(1)
(c) gradient $= \dfrac{3.0}{0.120}$ (or any two points read from the line) $= 25\ \mathrm{N\,m^{-1}}$ (accept 24–26 N m$^{-1}$; unit required)(1)
Answer: (a) $F = 2.0$ N point; exclude from line / repeat (b) uncertainty in $F$ negligible (c) 25 N m$^{-1}$
Examiner tip: Read the gradient from the drawn line using a large triangle, not from a single data point, and give the unit N m$^{-1}$.

5Standard6 marks

A student investigates how the resistance $R$ of a constantan wire depends on its length $L$. Two crocodile clips are attached to the wire and connected to an ohmmeter by leads. The diameter of the wire is 0.32 mm.

The graph shows the results. The line of best fit passes through (0, 0.30) and (1.00, 6.40).

(a) The theory predicts that $R = \dfrac{\rho L}{A}$. State what feature of the graph shows that there is a systematic error, and suggest a cause. [2]

(b) Determine the resistivity $\rho$ of constantan. [3]

(a) the line does not pass through the origin / has a non-zero intercept (of 0.30 Ω) on the $R$ axis(1)
resistance of the leads / of the crocodile clips / contact resistance (accept ohmmeter zero error; do not accept "human error" or "parallax")(1)
(b) gradient $= \dfrac{6.40 - 0.30}{1.00} = 6.1\ \Omega\,\mathrm{m^{-1}}$(1)
$A = \pi (0.16\times10^{-3})^{2} = 8.0\times10^{-8}\ \mathrm{m^{2}}$ (award if seen in working; using the diameter instead of the radius loses this mark)(1)
$\rho = \text{gradient}\times A = 4.9\times10^{-7}\ \Omega\,\mathrm{m}$ (ECF from gradient and area; unit required)(1)
(c) the gradient is unchanged because the error adds the same constant amount to every value of $R$ / it only shifts the line up (accept "only the intercept is affected")(1)
Answer: (a) non-zero intercept; lead/clip resistance (b) $4.9\times10^{-7}\ \Omega$ m (c) a constant offset does not change the gradient
Examiner tip: Use the gradient, not a single $R/L$ ratio: dividing one reading by its length includes the 0.30 Ω offset and gives a value that is too large.

6Standard6 marks

A student measures the period $T$ of oscillation of a mass $m$ on a spring. The student suggests that $T = c\,m^{n}$, where $c$ and $n$ are constants.

$m$ / kg	0.10	0.20	0.40	0.80
$T$ / s	0.50	0.70	0.99	1.41

(a) Explain why a graph of $\ln T$ against $\ln m$ can be used to find $n$. [2]

(b) The points lie on a straight line. Use the first and last data points to determine $n$. [2]

(a) taking logs gives $\ln T = n\ln m + \ln c$(1)
this has the form $y = mx + c$ so the graph is a straight line with gradient $n$ (and intercept $\ln c$)(1)
(b) gradient $= \dfrac{\ln 1.41 - \ln 0.50}{\ln 0.80 - \ln 0.10} = \dfrac{0.344-(-0.693)}{-0.223-(-2.303)}$(1)
$n = 0.50$ (accept 0.499; allow use of lg throughout)(1)
(c) $c = \dfrac{T}{\sqrt{m}} = \dfrac{1.41}{\sqrt{0.80}} = 1.58\ \mathrm{s\,kg^{-1/2}}$ OR $\ln c = 0.455$ (accept any data point: $c$ = 1.57–1.58)(1)
$k = \left(\dfrac{2\pi}{c}\right)^{2} = 16\ \mathrm{N\,m^{-1}}$ (accept 15.8–15.9 N m$^{-1}$)(1)
Answer: (a) $\ln T = n\ln m + \ln c$, gradient $= n$ (b) $n = 0.50$ (c) $k \approx 16$ N m$^{-1}$
Examiner tip: The intercept of a log–log graph is $\ln c$, not $c$; remember to take the exponential before using it.

7Standard4 marks

A rubber band is stretched by a force $F$. The graph shows how $F$ varies with the extension $x$. The data points are:

$x$ / m	0	0.02	0.04	0.06	0.08	0.10
$F$ / N	0	1.6	2.6	3.2	3.7	4.1

(a) State what is represented by the area under the graph. [1]

(b) Estimate the area under the graph between $x = 0$ and $x = 0.10$ m by treating each strip as a trapezium. [2]

(a) work done (on the band) in stretching it / elastic potential energy stored (allow energy transferred to the band)(1)
(b) sum of trapezia: $0.02\times\left(\dfrac{0+4.1}{2} + 1.6+2.6+3.2+3.7\right)$ OR five separate trapezium areas added(1)
$= 0.26\ \mathrm{J}$ (accept 0.263 J; accept 0.25–0.28 J from counting squares; unit required)(1)
(c) underestimate, because the curve bends downwards / is concave so the straight top edges of the trapezia lie below the curve(1)
Answer: (a) work done stretching the band (b) 0.26 J (c) underestimate
Examiner tip: A force–extension area is in N × m = J; for a curve, say which way the straight chords lie relative to the curve.

8Hard7 marks

An electric heater is placed in 0.250 kg of water in a polystyrene cup. The potential difference across the heater is 12.0 V and the current is 4.00 A. The temperature $\theta$ of the water is recorded every 30 s and a graph of $\theta$ against time $t$ is plotted.

The gradient of the line of best fit is $0.0440\ \mathrm{K\,s^{-1}}$. The steepest and shallowest lines that pass through all the error bars have gradients $0.0452\ \mathrm{K\,s^{-1}}$ and $0.0428\ \mathrm{K\,s^{-1}}$.

(a) Show that the specific heat capacity $c$ of water is given by $c = \dfrac{P}{m \times \text{gradient}}$, where $P$ is the power of the heater and $m$ is the mass of water. [1]

(b) Calculate $c$. [2]

(d) The accepted value of $c$ is $4180\ \mathrm{J\,kg^{-1}\,K^{-1}}$. Discuss, with reference to a systematic error, whether the student's result is consistent with this value. [2]

(a) $P\,\Delta t = mc\,\Delta\theta$ so $\dfrac{\Delta\theta}{\Delta t} = \dfrac{P}{mc}$ (= gradient) and rearranging gives the result(1)
(b) $P = 12.0\times4.00 = 48.0\ \mathrm{W}$(1)
$c = \dfrac{48.0}{0.250\times0.0440} = 4.36\times10^{3}\ \mathrm{J\,kg^{-1}\,K^{-1}}$ (accept 4360 or 4364)(1)
(c) $c_{\max} = \dfrac{48.0}{0.250\times0.0428} = 4486$ AND $c_{\min} = \dfrac{48.0}{0.250\times0.0452} = 4248$ OR percentage uncertainty in gradient $= \dfrac{0.0012}{0.0440} = 2.7\%$(1)
$\Delta c = \pm 1.2\times10^{2}\ \mathrm{J\,kg^{-1}\,K^{-1}}$ (accept $\pm$ 120 or $\pm$ 119; allow $\pm 1\times10^{2}$ to 1 s.f.; ignore extra precision; ECF from (b))(1)
(d) not consistent: 4180 lies outside the range 4240–4490 (allow range from candidate's (c))(1)
energy is lost to the surroundings / absorbed by the cup or heater, so the temperature rise is smaller, the gradient is too small and $c$ is too large (must link the error to the value being too high)(1)
Answer: (b) $4.36\times10^{3}$ J kg$^{-1}$ K$^{-1}$ (c) $\pm 1.2\times10^{2}$ J kg$^{-1}$ K$^{-1}$ (d) not consistent; heat loss makes $c$ too large
Examiner tip: When judging consistency, compare the accepted value with the range $c \pm \Delta c$, and make sure your systematic error pushes the result in the observed direction.

9Hard6 marks

A student drops a tennis ball from a height $H$ onto a hard floor and records the height $h$ of the first bounce by eye against a vertical metre rule. Each value of $H$ is used once. The uncertainty in $h$ is estimated as $\pm 0.02$ m.

$H$ / m	0.20	0.40	0.60	0.80	1.00	1.20
$h$ / m	0.15	0.29	0.42	0.53	0.62	0.70

The student draws the dashed straight line through the origin shown on the graph and concludes: "the bounce height is directly proportional to the drop height."

Evaluate the student's conclusion and the method used, and suggest improvements to the investigation. [6]

the data show a curve / the gradient decreases as $H$ increases, so the points do not lie on one straight line through the origin(1)
the dashed line misses the error bars of the last three points ($H$ = 0.80, 1.00 and 1.20 m), e.g. at $H$ = 1.20 m the line gives 0.84 m but $h$ = 0.70 $\pm$ 0.02 m; it only fits the first three(1)
$h/H$ falls from 0.75 to 0.58; even allowing for $\pm$0.02 m the ranges (0.65–0.85 and 0.57–0.60) do not overlap, so the trend is real(1)
conclusion is not supported over the whole range; proportionality may hold approximately only for small $H$ (first two or three points)(1)
possible physical reason: a greater fraction of energy is dissipated / air resistance is more significant at higher speeds(1)
method weaknesses: single readings so random error and anomalies cannot be spotted; judging the top of the bounce by eye gives parallax and reaction-time problems(1)
improvements: repeat each height several times and average; record with a video camera / phone at the level of the bounce and analyse frame by frame; use a fixed release mechanism; extend the range and include more values of $H$, especially at small $H$(1)
test a different model, e.g. plot $\ln h$ against $\ln H$ to find the power(1)

Level	Marks	What it looks like
Level 3	5–6	Uses the data quantitatively (e.g. ratio $h/H$ or comparison with error bars) to reject proportionality over the full range, identifies at least two weaknesses in the method and gives specific improvements linked to those weaknesses; reasoning is clear and logically ordered.
Level 2	3–4	Recognises that the data curve or do not fit the line, with some reference to the data or error bars; identifies at least one weakness and a matching improvement. Some links may be incomplete.
Level 1	1–2	Makes isolated relevant points, e.g. a generic improvement such as "repeat the readings" or a statement that the line is a poor fit, with little justification from the data.
Level 0	0	No relevant content.

Answer: Conclusion not supported: $h/H$ falls from 0.75 to 0.58, beyond the uncertainty; single readings judged by eye; repeat, use video analysis, more and lower values of $H$.
Examiner tip: "Evaluate" needs evidence from the data: quote values or error bars to show the trend is bigger than the uncertainty, then link each weakness to a specific improvement.

10StretchHL only6 marks

Monochromatic light of frequency $f$ falls on a metal surface and the stopping voltage $V_{s}$ is measured. The graph shows the results with error bars. The line of best fit passes through ($6.0\times10^{14}$ Hz, 0.20 V) and ($10.0\times10^{14}$ Hz, 1.86 V). Note that the horizontal axis does not start at zero.

(a) Using $eV_{s} = hf - \Phi$, determine a value for the Planck constant $h$ from the graph. [2]

(b) Determine the work function $\Phi$ of the metal, in eV. [2]

(c) The steepest and shallowest lines through the error bars have gradients $4.30\times10^{-15}$ V s and $4.00\times10^{-15}$ V s. Determine whether the experimental value of $h$ is consistent with the accepted value of $6.63\times10^{-34}$ J s. [2]

(a) gradient $= \dfrac{1.86-0.20}{(10.0-6.0)\times10^{14}} = 4.15\times10^{-15}\ \mathrm{V\,s}$(1)
$h = e\times\text{gradient} = 1.60\times10^{-19}\times4.15\times10^{-15} = 6.6\times10^{-34}\ \mathrm{J\,s}$ (accept 6.64)(1)
(b) threshold frequency $f_{0} = 6.0\times10^{14} - \dfrac{0.20}{4.15\times10^{-15}} = 5.5\times10^{14}\ \mathrm{Hz}$ (from extrapolating the line to $V_{s} = 0$) OR intercept on $V_{s}$ axis $= 0.20 - 4.15\times10^{-15}\times6.0\times10^{14} = -2.29$ V(1)
$\Phi = hf_{0}$ or $e\times|\text{intercept}| = 2.3\ \mathrm{eV}$ (accept 2.2–2.4 eV; ECF from (a); award [1 max] if the answer is left in joules: $3.7\times10^{-19}$ J)(1)
(c) $h$ range $= 1.60\times10^{-19}\times(4.00\text{ to }4.30)\times10^{-15} = 6.4\times10^{-34}$ to $6.9\times10^{-34}$ J s OR $h = (6.6 \pm 0.2)\times10^{-34}$ J s(1)
$6.63\times10^{-34}$ lies within this range so the value is consistent (conclusion must follow from a range)(1)
Answer: (a) $6.6\times10^{-34}$ J s (b) 2.3 eV ($f_0 = 5.5\times10^{14}$ Hz) (c) range $6.4$–$6.9\times10^{-34}$ J s contains 6.63, so consistent
Examiner tip: Because the horizontal axis starts at $5\times10^{14}$ Hz, the place where the line meets the vertical axis is not the true intercept; calculate it from $y = mx + c$.

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