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IB Physics – Data-Based Questions (Paper 1B)

Practice with unfamiliar experiments, uncertainties and graph analysis.

12 questions  ·  52 marks  ·  answers included  ·  IB Physics SL/HL (2023 syllabus)
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This worksheet covers:

  • This worksheet gives practice with unfamiliar experiments
  • Uncertainty propagation and graph analysis
  • In the style of IB Physics Paper 1B / data-based questions

A calculator and graph paper (or graphing software) are recommended; allow about 45 minutes.

Questions

Section A – Uncertainties (skills)

Calculator allowed
1

A student measures the length of a wire as $L = (25.0 \pm 0.1)\,\text{cm}$.Calculate the percentage uncertainty in $L$.

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$\dfrac{0.1}{25.0}\times 100\% = 0.4\%$
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2

Two lengths are measured as $l_1 = (12.4 \pm 0.2)\,\text{cm}$ and $l_2 = (8.6 \pm 0.1)\,\text{cm}$.(a) Calculate the sum $l_1 + l_2$ with its absolute uncertainty.(b) State the rule used for combining uncertainties in a sum.

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(a) $l_1+l_2 = (21.0 \pm 0.3)\,\text{cm}$(b) For a sum or difference, absolute uncertainties are added: $\Delta(l_1+l_2)=\Delta l_1+\Delta l_2$.
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3

A wire has diameter $d = (0.32 \pm 0.01)\,\text{mm}$ and length $L = (0.500 \pm 0.001)\,\text{m}$. Its resistance is measured as $R = (1.35 \pm 0.05)\,\Omega$.(a) Calculate the cross-sectional area $A$ of the wire.(b) Calculate the resistivity $\rho = \dfrac{RA}{L}$ of the wire.(c) Determine the percentage uncertainty, and hence the absolute uncertainty, in $\rho$.

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(a) $A = \pi (d/2)^2 = 8.04\times10^{-8}\,\text{m}^2$(b) $\rho = 2.17\times10^{-7}\,\Omega\,\text{m}$(c) $\frac{\Delta\rho}{\rho} = \frac{\Delta R}{R}+2\frac{\Delta d}{d}+\frac{\Delta L}{L} = 3.70\%+6.25\%+0.20\% = 10.2\%$, so $\Delta\rho \approx 2.2\times10^{-8}\,\Omega\,\text{m}$; $\rho = (2.17\pm0.22)\times10^{-7}\,\Omega\,\text{m}$
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4

A student uses a digital vernier caliper with a resolution of $0.01\,\text{mm}$ to take a single measurement of a rod’s diameter.State the absolute uncertainty that should be recorded, and explain why.

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$\pm 0.01\,\text{mm}$ (i.e. $\pm$ half of the smallest scale division, or the instrument’s stated resolution) – for a single reading this is the smallest interval the instrument can distinguish.
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Section B – Graph analysis (exam-style)

Calculator allowed
5

A student times $10$ oscillations of a simple pendulum of length $L = (0.850 \pm 0.002)\,\text{m}$ and finds the period $T = (1.851 \pm 0.006)\,\text{s}$.(a) Use $g = \dfrac{4\pi^2 L}{T^2}$ to calculate a value for $g$.(b) Calculate the percentage uncertainty in $g$.(c) State the absolute uncertainty in $g$, and hence give $g$ with its uncertainty to an appropriate number of significant figures.

Show answer
(a) $g = 9.79\,\text{m s}^{-2}$(b) $\frac{\Delta g}{g} = \frac{\Delta L}{L}+2\frac{\Delta T}{T} = 0.24\%+0.65\% = 0.88\%$(c) $\Delta g \approx 0.09\,\text{m s}^{-2}$, so $g = (9.79 \pm 0.09)\,\text{m s}^{-2}$
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6

Gamma radiation passes through absorbers of thickness $x$ and the intensity $I$ is recorded:$x/\text{cm}$: 0, 2, 4, 6, 8, 10$I/\text{counts s}^{-1}$: 820, 610, 450, 335, 250, 185The intensity follows $I = I_0 e^{-\mu x}$.(a) Explain why plotting $\ln I$ against $x$ should give a straight line, and state the significance of the gradient and the $y$-intercept.(b) Given that the gradient of the best-fit line of $\ln I$ against $x$ is $-0.149\,\text{cm}^{-1}$ and the $y$-intercept is $6.71$, find the attenuation coefficient $\mu$ and the value of $I_0$.(c) Calculate the half-value thickness of the absorber.

Show answer
(a) Taking logs: $\ln I = \ln I_0 – \mu x$, which has the form $y=mx+c$ with $y=\ln I$, $x=x$, gradient $m=-\mu$ and intercept $c=\ln I_0$.(b) $\mu = 0.149\,\text{cm}^{-1}$; $I_0 = e^{6.71} = 820\,\text{counts s}^{-1}$(c) $x_{1/2} = \dfrac{\ln 2}{\mu} = 4.66\,\text{cm}$
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7

In a timing experiment a student measures the period of a single oscillation using a hand-held stopwatch.Suggest one change to the method that would reduce the random uncertainty in the period, and explain why it works.

Show answer
Time a large number of oscillations (e.g. 20) and divide the total time by that number: the fixed reaction-time uncertainty in starting/stopping the stopwatch is then divided by the number of oscillations, reducing the uncertainty in a single period.
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8

A student measuring the resistivity of a wire notices that every resistance reading is $0.3\,\Omega$ higher than expected because the ammeter was not zeroed before use.(a) State whether this is a random or a systematic error.(b) Explain the effect this error has on the calculated resistivity, and how the error could be removed.

Show answer
(a) Systematic error.(b) It shifts every resistance value up by the same fixed amount, so the calculated resistivity is too high by a constant fraction; it does not increase the scatter/uncertainty. It can be removed by zeroing (checking for zero error on) the ammeter before taking readings, or by subtracting the known offset from every reading.
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Section C – Extended investigation (challenge)

Calculator allowed
9

Small steel spheres of radius $r$ fall at terminal velocity $v$ through a viscous liquid. Data collected:$r/\text{mm}$: 0.5, 1.0, 1.5, 2.0$v/\text{cm s}^{-1}$: 2.1, 8.3, 18.9, 33.5(a) Show that these data are consistent with $v \propto r^2$.(b) The relationship is $v = \dfrac{2r^2(\rho_s-\rho_f)g}{9\eta}$, where $\rho_s = 2500\,\text{kg m}^{-3}$ and $\rho_f = 960\,\text{kg m}^{-3}$. A graph of $v$ against $r^2$ has gradient $8.4\times10^4\,\text{s}^{-1}\,\text{m}^{-1}$ (SI units). Calculate the viscosity $\eta$ of the liquid.

Show answer
(a) $v/r^2$ is constant ($8.4, 8.3, 8.4, 8.4\,\text{cm s}^{-1}\text{mm}^{-2}$ to 2 s.f.), confirming $v \propto r^2$.(b) $\eta = \dfrac{2(\rho_s-\rho_f)g}{9\times(\text{gradient})} = 4.0\times10^{-2}\,\text{Pa s}$
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10

In the pendulum experiment of Question 5, the accepted value of $g$ at the location is $9.81\,\text{m s}^{-2}$.(a) Calculate the percentage difference between the experimental value and the accepted value.(b) State, with a reason, whether the experimental value is consistent with the accepted value.

Show answer
(a) $\dfrac{|9.79-9.81|}{9.81}\times100\% = 0.2\%$(b) Yes – the percentage difference (0.2%) is smaller than the percentage uncertainty in the experimental value (0.9%), so the accepted value lies within the experimental uncertainty range $(9.79\pm0.09)\,\text{m s}^{-2}$.
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11

A student investigates how the period $T$ of a pendulum depends on its length $L$, believing $T = kL^{n}$.$L/\text{m}$: 0.20, 0.40, 0.60, 0.80, 1.00$T/\text{s}$: 0.90, 1.27, 1.55, 1.79, 2.01(a) Explain how plotting $\log T$ against $\log L$ allows $n$ and $k$ to be determined.(b) The best-fit line through $\log T$ vs $\log L$ has gradient $0.498$ and $y$-intercept $0.302$. Determine $n$ and $k$, giving each to an appropriate number of significant figures.(c) State the physical significance of the value of $n$ found.

Show answer
(a) Taking logs of $T=kL^n$ gives $\log T = n\log L + \log k$, of the form $y=mx+c$: the gradient equals the power $n$ and the intercept equals $\log k$, so both can be read from a straight-line graph.(b) $n = 0.498 \approx 0.50$; $k = 10^{0.302} = 2.00$(c) $n \approx \dfrac{1}{2}$ confirms $T \propto \sqrt{L}$, consistent with $T = 2\pi\sqrt{L/g}$.
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12

For the gamma-ray absorption experiment in Question 6, suggest one improvement to the experimental method that would reduce the uncertainty in the measured count rate, and justify your answer.

Show answer
Take counts over a longer time interval (or repeat and average several counts) at each thickness, and subtract the background count rate measured with no absorber present: radioactive decay is random, so longer counting times and background subtraction reduce the statistical (Poisson) uncertainty and remove a systematic offset from background radiation.
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