2025 OCR A-Level Physics A - H556 模拟试题及答案详解

\(R = V/I = 6.0/0.50 = 12\ \Omega\). Percentage uncertainties: V: \(0.1/6.0 = 1.67\%\); I: \(0.01/0.50 = 2.0\%\). Add: 3.67%. Absolute uncertainty \(= 0.0367 \times 12 = 0.44 \approx 0.4\ \Omega\). Answer A.

A (1 mark). B1 for A only. Common errors: B adds absolute uncertainties (0.11); C uses only the p.d. percentage (0.2); D adds percentages wrongly/ignores division.

\(F = BIL\sin\theta = 0.040 \times 3.0 \times 0.25 \times \sin 30^\circ = 0.015\) N. Answer B.

B (1 mark). A uses cos 30°; C omits the sine; D applies the sine factor twice.

Energy \(= Pt = 40 \times 150 = 6000\) J. \(c = E/(m\Delta\theta) = 6000/(0.50 \times 12) = 1000\ \mathrm{J\,kg^{-1}\,K^{-1}}\). Answer D.

D (1 mark). C omits the mass (6000/12); B divides by 0.50 and 12 and a further 2; A divides 1000 by 12 again.

24 mm is three half-value thicknesses, so \(I/I_0 = (1/2)^3 = 1/8 = 12.5\%\). Answer B.

B (1 mark). A uses two half-value thicknesses; C and D come from treating the attenuation as linear.

6.0 h is three half-lives: \(800 \to 400 \to 200 \to 100\) Bq. Answer A.

A (1 mark). B is two half-lives; C assumes linear decay; D is one half-life.

Path difference \(= 4.5/3.0 = 1.5\lambda\), an odd number of half-wavelengths, so destructive interference (minimum). Phase difference \(= 1.5 \times 2\pi = 3\pi\) rad. Answer C.

C (1 mark). Distractors misidentify the interference type or use \(\pi\) per wavelength incorrectly.

\(V = -GM/r\). At distance \(2R\), \(V = -3.0 \times 10^7\ \mathrm{J\,kg^{-1}}\). \(\Delta V = 3.0 \times 10^7\). Work \(= m\Delta V = 2.0 \times 3.0 \times 10^7 = 6.0 \times 10^7\) J. Answer B.

B (1 mark). A omits the mass; C is the work to reach infinity; D uses a wrong potential ratio.

Vertical speed: \(v_y^2 = 2gh = 2 \times 9.81 \times 10\), so \(v_y = 14.0\ \mathrm{m\,s^{-1}}\). Horizontal speed stays 15. \(v = \sqrt{15^2 + 14.0^2} = 20.5 \approx 21\ \mathrm{m\,s^{-1}}\). Answer C.

C (1 mark). A gives only the vertical component; B only the horizontal; D adds the components arithmetically.

\(R = V/I = 6.0/0.50 = 12\ \Omega\). Percentage uncertainties: \(0.2/6.0 = 3.3\%\) and \(0.02/0.50 = 4.0\%\). Add for a quotient: \(7.3\%\). Absolute uncertainty \(= 0.073 \times 12 \approx 0.9\ \Omega\). Answer C.

C (1 mark). B1 for the correct option only. Guidance: A arises from using only the voltage fractional uncertainty incorrectly; B from using a single fractional uncertainty; D from adding absolute uncertainties incorrectly.

\(800 \to 400 \to 200 \to 100\) requires three half-lives. \(t = 3 \times 6.0 = 18\) hours. Answer B.

B (1 mark). B1 for the correct option only. Guidance: A is two half-lives; C is four half-lives; D is an incorrect count (six half-lives).

Vertical motion: \(20 = \tfrac{1}{2} g t^2\), so \(t = \sqrt{2 \times 20 / 9.81} = 2.02\) s. Horizontal distance \(= 12 \times 2.02 \approx 24\) m. Answer B.

B (1 mark). B1 for the correct option only. Guidance: the working is M1 for time of fall from vertical motion, A1 for distance, but only the letter is credited.

Energy required \(= m \Delta V = 50 \times (0 - (-6.4 \times 10^{7})) = 3.2 \times 10^{9}\) J. Answer D.

D (1 mark). B1 for the correct option only. Guidance: A has the wrong sign; B divides instead of multiplying; C omits the mass.

\(\lambda = a x / D = (0.50 \times 10^{-3} \times 2.4 \times 10^{-3}) / 2.0 = 6.0 \times 10^{-7}\) m. Answer A.

A (1 mark). B1 for the correct option only. Guidance: B arises from failing to convert millimetres to metres; C from inverting the relationship.

\(\Delta\lambda / \lambda = v / c\), so \(v = (5/500) \times 3.00 \times 10^{8} = 3.0 \times 10^{6}\ \text{m s}^{-1}\). Answer D.

D (1 mark). B1 for the correct option only. Guidance: C omits the fractional shift; A and B are factor-of-two errors.

\(F = B I L \sin\theta = 0.20 \times 5.0 \times 0.40 \times \sin 30^\circ = 0.20\) N. Answer C.

C (1 mark). B1 for the correct option only. Guidance: B uses cosine instead of sine; D omits the angle; A is a factor-of-two error.

(a) \(0.35 = e^{-\mu \times 2.0}\), so \(\ln 0.35 = -2.0\mu\). \(\mu = 1.0498/2.0 = 0.525\ \text{mm}^{-1}\).

(b) \(0.10 = e^{-0.525x}\), so \(x = \ln 10 / 0.525 = 2.303/0.525 = 4.4\ \text{mm}\) (4.39).

(c) Low-energy photons have a larger attenuation coefficient, so they are absorbed in the soft tissue of the patient. They add to the dose but are not transmitted to the detector, so they do not contribute to the image. The filter removes them from the beam, reducing the dose, and the remaining higher-energy beam gives a sharper image with better contrast between bone and tissue.

(a) M1: rearranges to \(\mu = -\ln(0.35)/2.0\) (or equivalent use of ln). A1: \(\mu = 0.52\) or \(0.525\ \text{mm}^{-1}\). Allow 2 or more s.f. Allow \(525\ \text{m}^{-1}\) if converted correctly.

(b) M1: \(x = \ln(10)/\mu\) (or \(\ln(1/0.10)/\mu\)). A1: \(4.4\ \text{mm}\) (accept 4.38 to 4.40, AW). ECF from (a) for the A mark. Allow a correct alternative method using (a) and the ratio of ln values.

(c) B1: low-energy (soft) X-ray photons are absorbed more readily / strongly attenuated by the filter (or by the patient's tissue). B1: these photons would be absorbed in the patient, adding to the dose without reaching the detector / contributing to the image, so removing them reduces dose and improves contrast or sharpness. Allow: beam becomes more penetrating / more monochromatic. Ignore: statements that the filter reduces intensity only. NOT: the filter absorbs all photons equally.

(a)(i) \(I = I_0 e^{-\mu x}\), so \(210 = 840\,e^{-\mu(0.0040)}\). Then \(e^{-\mu x}=0.25\), so \(\mu = \ln 4/0.0040 = 1.386/0.0040 = 347\ \text{m}^{-1}\), which is approximately \(350\ \text{m}^{-1}\).
(a)(ii) \(I/I_0 = 0.050\), so \(\mu x = \ln 20 = 3.00\). \(x = 3.00/347 = 8.6\times10^{-3}\ \text{m}\) (8.6 mm).
(b) \(x_{1/2} = \ln 2/\mu = 0.693/347 = 2.0\times10^{-3}\ \text{m}\) (2.0 mm). This is consistent with 4.0 mm being two half-value thicknesses (840 to 420 to 210).
(c) Low-energy photons are strongly attenuated by the body (large \(\mu\)), so they are absorbed in the skin and tissue near the surface and add to the dose without reaching the detector. The aluminium filter removes a large fraction of these low-energy photons. Higher-energy photons are attenuated much less, so they pass through the filter and the patient to form the image, and image quality is little affected.

(a)(i) M1: correct substitution into \(I = I_0 e^{-\mu x}\) with ratio 0.25 (or 4) and ln taken [1]. A1: \(\mu = 347\ \text{m}^{-1}\) shown, with an unrounded value (or 3 s.f.) seen before the final statement [1]. Allow 0.347 mm^-1 or 0.347 per mm only with consistent units. Allow the half-value method: two half-value thicknesses so \(x_{1/2} = 2.0\) mm, then \(\mu = \ln 2/x_{1/2}\) (M1), \(= 347\ \text{m}^{-1}\) (A1).
(a)(ii) C1/M1: \(\ln(0.050)=-\mu x\) or \(\ln 20 = \mu x\) [1]. A1: \(x = 8.6\times10^{-3}\ \text{m}\) (8.6 mm) [1]. Allow 8.6 to 8.7 mm; ECF from a wrong \(\mu\) in (a)(i) only if \(\mu\) was calculated, not the given 350. Using \(\mu=350\) gives 8.56 mm: allow. NOT 5.0 as a fraction used in place of 0.050 (gives zero for the second mark).
(b) B1: 2.0 mm (\(2.0\times10^{-3}\) m), from \(\ln 2/\mu\) or from 4.0 mm / 2. Allow 1.98 to 2.0 mm. ECF from \(\mu\) in (a)(i). [1]
(c) B1: low-energy (soft) X-ray photons are absorbed by the filter / are preferentially attenuated (AW higher attenuation coefficient) [1]. B1: these photons would be absorbed by the patient (skin) and so increase dose without contributing to the image; high-energy photons pass through to form the image [1]. Allow ora: the filter hardens the beam. Ignore reference to the wavelength of the X-rays unless connected to energy. NOT 'the filter blocks all radiation' or 'filter removes harmful radiation' without reference to energy.

(a) \(\mu x = 140 \times 4.0\times10^{-3} = 0.56\). \(\dfrac{I}{I_0} = e^{-0.56} = 0.571 \approx 0.57\).

(b) Measure the background count/intensity with the source blocked and subtract it. Measure the thickness of a sheet with a micrometer, then record the transmitted intensity \(I\) for increasing numbers of sheets (thickness \(x\)), keeping the source-detector distance fixed and using a narrow collimated beam. Repeat each reading and average. Taking natural logs gives \(\ln I = \ln I_0 - \mu x\). Plot \(\ln I\) against \(x\): a straight line with negative gradient, so \(\mu = -\text{gradient}\). Safety: shield the source with lead, keep exposure time short, stay at a distance or behind a screen, and do not stand in the beam. Uncertainty: random variation in the detector reading, or the uncertainty in the sheet thickness, or sheets not exactly identical.

(a) M1: \(\mu x = 140 \times 4.0\times10^{-3} = 0.56\) (unit conversion from mm to m must be correct). A1: \(I/I_0 = e^{-0.56} = 0.57\) (0.571). Allow 57% or answers to 2 s.f. or more. ECF from an incorrect \(\mu x\) for A1 only if the exponential is evaluated correctly. Answer of 0.57 with no working gets 2.

(b)* Level of response, assessed holistically (6 marks).

Band 3 (5-6 marks): A complete, logically structured method which includes measurements, repeated/controlled procedure, a correct linearised graph with the gradient linked to \(\mu\), and both a safety precaution and a source of uncertainty. The reasoning is clear and well supported by physics, with at most minor omissions.

Band 2 (3-4 marks): A method covering most strands, e.g. the measurement of I with varying thickness and a mostly correct graph, but one strand is missing or weak (e.g. no mention of safety or uncertainty, or no control of variables, or the graph relationship is not explained fully).

Band 1 (1-2 marks): Some relevant points made (e.g. measure I for different thicknesses, or plot a graph) but the method is incomplete, or the analysis is not linked to \(I = I_0e^{-\mu x}\). Little structure.

0 marks: No creditworthy response.

Indicative scientific points (not a compulsory script):
- measure background and subtract (correct for background)
- measure thickness of sheets with micrometer / vernier calipers; use stack of n sheets
- measure transmitted intensity I for a range of thicknesses x, including x = 0 for I0
- fixed source-detector distance, same beam alignment / narrow collimated beam / same source settings
- repeat readings and average
- \(\ln I = \ln I_0 - \mu x\)
- plot \(\ln I\) (or \(\ln(I/I_0)\)) on the y-axis against x on the x-axis
- straight line with negative gradient; \(\mu = -\) gradient (large triangle)
- safety: lead shielding, minimise exposure time, keep distance, do not stand in beam, warning signs/ trained operator
- uncertainty: random fluctuation in readings, thickness measurement, non-identical sheets, detector resolution, beam not perfectly narrow

Allow lg instead of ln only if the gradient is correctly converted (\(\mu = -2.303\times\) gradient). NOT a graph of I against x with gradient equal to \(\mu\). Ignore reference to half-value thickness unless the method is otherwise correct.

(a)(i) \(I/I_0 = e^{-\mu x} = e^{-0.28\times 6.0} = e^{-1.68} = 0.186 \approx 0.19\).

(a)(ii) At the half-value thickness \(I/I_0 = 0.5\), so \(\ln 2 = \mu x_{1/2}\). Then \(x_{1/2} = 0.693/0.28 = 2.48\ \text{cm} \approx 2.5\ \text{cm}\).

(b) Bone has a larger attenuation coefficient than soft tissue (higher density and proton number), so it attenuates the beam more strongly by the photoelectric effect (also Compton scattering). The transmitted intensity behind bone is therefore lower than behind soft tissue, giving contrast between the two regions of the image. The aluminium sheet absorbs low-energy (soft) X-ray photons, which are strongly attenuated and would be absorbed in the patient's tissue without reaching the detector. Removing them reduces the dose to the patient while leaving the higher-energy photons that form the image. The beam becomes more penetrating (harder) and more nearly of one energy, which gives a more consistent image, although the overall intensity at the detector is reduced slightly.

(a)(i) M1: substitution \(e^{-0.28\times6.0}\) or \(e^{-1.68}\) [1]. A1: 0.19 (allow 0.186; 19% accept) [1]. Guidance: answer of 5.4 to 5.5 (inverse) scores M1 only; do not penalise 2 s.f. or 3 s.f.

(a)(ii) B1: \(0.5 = e^{-\mu x}\) so \(x = \ln 2/\mu\), and evaluates \(0.693/0.28 = 2.48\) (cm) [1]. Guidance: the working must show the calculated value to at least 2 s.f. before rounding to 2.5.

(b)* Level of response, 6 marks.
Indicative scientific points:
- Bone has a greater attenuation coefficient / is denser / has a higher proton number than soft tissue (AW).
- Bone absorbs or scatters more X-ray photons so the transmitted intensity is lower behind bone (ORA); regions of different intensity give contrast (bone appears lighter on the radiograph/lower exposure on detector, AW).
- Named attenuation process: photoelectric effect, Compton scattering, (coherent/simple scattering or pair production, the latter only at much higher photon energies). Allow any one correctly named process.
- The aluminium sheet (filter) preferentially removes low-energy photons (since attenuation is greater for lower photon energy).
- Low-energy photons would be absorbed by the patient's tissue and not contribute to the image, so the filter reduces patient dose.
- The remaining beam has a higher mean photon energy / is harder / more penetrating; image quality is maintained or improved (a small reduction in total intensity may be mentioned).

5-6 marks: Relevant, well-supported reasoning covering all three bullet strands. Links the differing attenuation of bone and tissue to intensity differences and contrast, names a process correctly, and explains the filtering of low-energy photons with its effect on dose. The answer is logically structured and complete.
3-4 marks: Addresses at least two strands with some correct reasoning, but one is incomplete, or links between ideas are limited (e.g. states that the filter reduces dose without saying why, or states bone absorbs more without relating to intensity or contrast).
1-2 marks: Isolated correct statements from the indicative content, with little or no development or linking.
0 marks: No creditworthy response.
Guidance: Ignore reference to the film being darker/lighter provided contrast is explained consistently. NOT 'bone blocks all X-rays'. NOT 'filter removes all low-energy photons' without qualification (AW partial). Do not credit ionisation of the aluminium as the reason for the filter.

(a) Set \(I = I_0/2\): \(\tfrac{1}{2} = e^{-\mu x_{1/2}}\). Taking natural logarithms: \(\ln(1/2) = -\mu x_{1/2}\), so \(-\ln 2 = -\mu x_{1/2}\) and \(x_{1/2} = \ln 2/\mu\).

(b) \(I/I_0 = 0.050 = e^{-\mu x}\), so \(x = \ln(1/0.050)/\mu = \ln 20 / 0.80 = 2.996/0.80 = 3.74\,\text{cm} \approx 3.7\,\text{cm}\).

(c) Path 1: \(I_1 = I_0 e^{-0.18\times 4.0} = I_0 e^{-0.72}\). Path 2: 3.0 cm soft tissue and 1.0 cm bone: \(I_2 = I_0 e^{-(0.18\times 3.0 + 0.62\times 1.0)} = I_0 e^{-1.16}\). Ratio \(I_2/I_1 = e^{-0.44} = 0.64\). So the intensity through bone is only about 36% lower than through soft tissue alone, so contrast is modest: bone is distinguishable but not strongly. Contrast is the difference in degree of blackening between adjacent regions and depends on the difference in attenuation coefficients. Soft tissues have very similar \(\mu\), so little contrast; a contrast medium of high atomic number (e.g. barium) has a much larger \(\mu\) and so strongly attenuates X-rays, making the organ stand out. Sharpness (clarity of edges) is improved by a small X-ray source/narrow, parallel beam, patient close to the detector (less geometric blurring/penumbra), a thin intensifying screen or small detector pixels, and a grid reducing scattered X-rays; filtering/low-energy removal and a short exposure also reduce unwanted dose and movement blur.

(a) B1: \(\tfrac{1}{2} I_0 = I_0 e^{-\mu x_{1/2}}\) (or \(I/I_0 = 1/2\) substituted). B1: takes natural logs, \(\ln(1/2) = -\ln 2\), to reach \(x_{1/2} = \ln 2/\mu\). Guidance: the second mark needs a visible log step; do not allow a bare statement of the final expression.

(b) M1: \(0.050 = e^{-0.80x}\) or \(x = \ln(20)/0.80\) (allow \(\ln 0.05 = -0.80x\)). A1: \(x = 3.7\,\text{cm}\) (allow 3.74; allow 3.8 only if 3.75 shown). Guidance: a final answer of 3.7 with no working scores 2. Allow 0.037 m. ECF not applicable.

(c)* Level of response, assessed holistically.

Indicative scientific points:
- Correct calculation: \(I_2/I_1 = e^{-(0.62-0.18)\times 1.0} = e^{-0.44} = 0.64\) (allow 0.64–0.65; the inverse 1.55 is accepted if clearly labelled as path 1 / path 2). Allow ECF into the interpretation.
- Interpretation: contrast is the difference in blackening/intensity between regions, dependent on the difference in \(\mu\); a ratio of 0.64 means moderate contrast, only about 36% less intensity through bone.
- Soft tissues have very similar \(\mu\), so adjacent soft organs give poor contrast.
- A contrast medium (e.g. barium or iodine compound, high atomic number) has a much larger \(\mu\) (greater absorption, mainly by the photoelectric effect), so the organ is outlined clearly.
- Sharpness: clear edges. Improved by small source or narrow beam, patient close to detector/film, thin or fine-grain intensifying screen or small pixel size, anti-scatter grid, short exposure to avoid movement blur.
- Allow reference to scattering reducing contrast and sharpness.

5–6 marks: calculation present and correct (or only one slip), correctly interpreted for contrast, AND both the contrast-medium and sharpness strands covered with correct physics, linked and logically structured. Six marks requires at least two valid sharpness measures with reasons and a clear link between \(\mu\) values and the contrast.
3–4 marks: calculation attempted with correct method (may contain an arithmetic slip) and at least two of the three strands addressed with some correct physics, but reasoning is incomplete or one strand is only stated without explanation.
1–2 marks: isolated correct points (e.g. a contrast medium absorbs more; small source gives a sharper image; correct use of the attenuation equation for one path) without a coherent line of reasoning.
0 marks: no creditworthy response.

Ignore: references to dose or hazards unless linked to image quality. NOT: 'contrast medium makes X-rays pass through'. AW for sensible alternatives. Total 10 marks.

(a) \(I/I_0 = 217/620 = 0.35\). \(\ln(0.35) = -\mu \times 6.0\), so \(\mu = 1.0498/6.0 = 0.175\ \mathrm{mm^{-1}}\).

(b) \(0.050 = e^{-\mu x}\), so \(x = \ln(20)/\mu = 2.996/0.175 = 17.1\ \mathrm{mm}\) (about 17 mm).

(c) At the half-value thickness \(I = I_0/2\), so \(x_{1/2} = \ln 2/\mu = 0.693/0.175 = 3.96\ \mathrm{mm}\).

(d) \(\mu = \ln(I_0/I)/x\), so the unit is \(\mathrm{m^{-1}}\).

(e) X-rays from a small source pass through the arm and are attenuated exponentially. Bone has a higher density and a higher proton number than soft tissue, so it has a larger \(\mu\) and absorbs more X-rays. Fewer X-rays reach the detector behind bone, so bone appears lighter on the film (shadow of the transmitted intensity), with the fracture seen as a gap or line. Sharpness is improved by using a small point source, a narrow collimated beam, placing the detector close to the patient, and a lead grid to absorb scattered X-rays. Barium has a high proton number, so it has a large attenuation coefficient. The soft stomach tissue has attenuation similar to its surroundings, so barium increases the contrast and makes the outline of the stomach visible. Barium sulfate is insoluble and so is not absorbed into the body.

(a) M1: \(\ln(217/620)\) or \(\ln(0.35) = -\mu \times 6.0\) (or \(e^{-6\mu}=0.35\)). A1: \(\mu = 0.175\ \mathrm{mm^{-1}}\) with at least 3 s.f. shown (e.g. 0.1750 or 0.17497). Allow ECF only for correctly rearranged working. [2]

(b) M1: \(\ln(1/0.05)=\ln 20\) (= 2.996) divided by \(\mu\), or \(0.05 = e^{-0.175x}\). A1: 17 mm (17.1 mm). Allow 17.0 to 17.2. Allow answer in m (0.017 m). [2]

(c) M1: \(x_{1/2} = \ln 2/\mu\) (or \(0.5 = e^{-\mu x}\)). A1: 3.96 mm. Allow 4.0 mm. Allow ECF from (a) if a different \(\mu\) was used. [2]

(d) B1: \(\mathrm{m^{-1}}\). Allow 1/m. NOT \(\mathrm{mm^{-1}}\) or \(\mathrm{cm^{-1}}\). [1]

(e)* Level of response (6 marks). Assess holistically.
Band 3 (5-6 marks): all three strands (image formation including bone vs soft tissue; sharpness; barium meal) are covered with correct, clearly linked physics. Explanations include the reasons (attenuation depends on density and proton number; contrast arises from different attenuation). The line of reasoning is coherent and well structured.
Band 2 (3-4 marks): two strands are covered well, or all three covered with limited explanation or one significant omission/error. Reasoning is mostly logical.
Band 1 (1-2 marks): some relevant points stated, with little or no explanation, or only one strand addressed.
Band 0 (0 marks): no creditworthy response.

Indicative scientific points:
- X-rays are produced by a small source and pass through the body; the detector (film/digital sensor) records transmitted intensity.
- Different tissues attenuate X-rays differently; \(I = I_0 e^{-\mu x}\).
- Bone has larger density / higher proton number (Z), so larger \(\mu\); more absorbed, so fewer X-rays reach the detector (appears lighter/whiter on a conventional film; a fracture shows as a darker line or gap).
- Soft tissue absorbs less, so more X-rays are transmitted.
- Sharpness: small (point) source / small focal spot; collimator for a narrow parallel beam; detector close to the patient to reduce geometric blurring/penumbra; patient still; anti-scatter (lead) grid; intensifying screen allows lower dose.
- Barium meal: barium has a high proton number / high \(\mu\); stomach is soft tissue with similar attenuation to surroundings so there would be poor contrast; barium absorbs strongly, so the stomach's outline/lining is seen. Allow: barium is non-toxic as insoluble sulfate.

Allow: AW for all points. Accept 'contrast' for 'difference in shade'. Ignore: references to CT scans or ultrasound. NOT: bone 'reflects' X-rays; NOT: 'X-rays cannot pass through bone' (absorbed more, not blocked totally).

(a) The magnitude of the induced e.m.f. is equal (directly proportional) to the rate of change of magnetic flux linkage; the sign is such that it opposes the change (Lenz's law).

(b)(i) \(N\Phi = BAN = 0.35 \times 4.0\times10^{-4} \times 200 = 0.028\ \mathrm{Wb\ turns}\).
(b)(ii) Final flux linkage is zero, so change = \(0.028\). \(\varepsilon = 0.028 / 0.040 = 0.70\ \mathrm{V}\).

(c) \(\omega = 2\pi f = 2\pi \times 25 = 157\ \mathrm{rad\ s^{-1}}\). Maximum e.m.f. occurs when \(\sin\omega t = 1\): \(\varepsilon_{max} = BAN\omega = 0.028 \times 157 = 4.4\ \mathrm{V}\).

(d) An alternating current in the primary coil produces an alternating magnetic flux in the soft-iron core. This flux links the secondary coil, so a changing flux linkage induces an alternating e.m.f. in the secondary. Since \(V_s/V_p = N_s/N_p\), fewer turns on the secondary than on the primary gives a lower secondary p.d. The core is soft iron because it is easily magnetised and demagnetised, so little energy is lost in repeatedly reversing its magnetisation (hysteresis) and it concentrates the flux so that nearly all of it links the secondary. Lamination (thin insulated sheets) increases the resistance of paths for eddy currents, which reduces the currents induced in the core and so reduces heating. Real transformers are less than 100% efficient because of resistive heating (\(I^2R\)) in the windings, eddy currents and hysteresis in the core, and flux leakage so that not all primary flux links the secondary.

(a) B1: e.m.f. (induced) is proportional / equal to rate of change. B1: of (magnetic) flux linkage. Allow 'rate of cutting of flux' for B1 only if 'linkage' not stated. Allow Lenz's law statement about opposing the change as an alternative B1 only if the first B1 is also awarded. NOT 'change of flux' without 'rate'.

(b)(i) B1: 0.028 (Wb turns) (accept 2.8 x 10^-2). Allow 0.028 with unit Wb or Wb turns; NOT missing unit if wrong unit given.
(b)(ii) M1: \(\varepsilon = \Delta(N\Phi)/\Delta t\) with \(\Delta (N\Phi)=0.028\) and \(\Delta t = 0.040\). A1: 0.70 V. ECF from (b)(i). Allow 0.7 V. Ignore sign.

(c) B1: \(\omega = 2\pi \times 25 = 157\ \mathrm{rad\ s^{-1}}\) (or 160 to 2 s.f.). M1: \(\varepsilon_{max} = BAN\omega\) (sin = 1). A1: 4.4 V (accept 4.40, 4.5 from 160). ECF from (b)(i) value for BAN. NOT use of f in place of omega (gives 0.70 V: max 1 mark).

(d)* Level 3 (5-6 marks): all three bullet points are addressed with clear, correct, linked reasoning, including the role of changing flux in the core linking the secondary and the turns ratio, plus at least two valid energy-loss mechanisms explained. Level 2 (3-4 marks): two bullet points addressed with some correct reasoning, or all three addressed with limited explanation or minor errors. Level 1 (1-2 marks): some relevant correct points but reasoning is limited or incomplete. 0 marks: no creditworthy response.
Indicative scientific points: alternating current in primary produces changing magnetic flux in the core; core links flux to secondary; changing flux linkage induces e.m.f. in the secondary (Faraday); \(V_s/V_p = N_s/N_p\) so fewer secondary turns gives lower p.d.; soft iron is easily magnetised and demagnetised / low hysteresis loss; soft iron has high permeability so flux is concentrated in the core; lamination increases resistance to eddy currents; reduces eddy current heating; real losses: resistance of windings (I^2R heating); eddy currents; hysteresis; flux leakage / not all flux links the secondary. Allow 'energy lost as heat/sound' for losses if linked to a cause. Ignore references to the transformer producing direct current. NOT 'transformer creates energy'.

(a)(i) \(P = VI = 12.0 \times 4.00 = 48.0\) W.
(ii) \(\%V = 0.1/12.0 = 0.83\%\); \(\%I = 0.05/4.00 = 1.25\%\). Sum = 2.1%.
(b) Gradient \(= (35.7-20.0)/300 = 15.7/300 = 0.0523\) K s\(^{-1}\).
(c) \(P = mc\,\Delta\theta/\Delta t\), so \(c = P/(m \times \text{gradient}) = 48.0/(1.00 \times 0.05233) = 917\) J kg\(^{-1}\) K\(^{-1}\).
(d) Total percentage uncertainty = 1% (mass) + 2.08% (power) + 3% (gradient) = 6.1%. Absolute uncertainty = 0.061 × 917 = 56 J kg\(^{-1}\) K\(^{-1}\). So \(c = (920 \pm 60)\) J kg\(^{-1}\) K\(^{-1}\).
(e) The range is 861 to 973 J kg\(^{-1}\) K\(^{-1}\). 897 lies inside it, so the result is consistent with the accepted value. The measured value is slightly high, which is consistent with a small energy loss, but the uncertainty is too large to confirm a systematic error.
(f) Energy lost means the block's temperature rises less than it would without losses, so the gradient is smaller and the calculated \(c\) is too large (the heater energy is assumed to all heat the block). Losses increase as the block gets hotter, so the graph curves and the gradient falls with time; a straight line suggests losses are small. Improvements: more or thicker insulation, a lid; add oil or conducting paste in the thermometer and heater holes for good thermal contact; start the block a few degrees below room temperature so the average temperature is close to room temperature; use a smaller temperature range or repeat the experiment and average.

(a)(i) 48.0 W [B1]. Allow 48 W.
(a)(ii) Percentage uncertainties in V and I found: 0.83% and 1.25% [M1]; added to give 2.1% [A1]. Allow 2.08% or 2%. Allow 1 mark for an absolute uncertainty of 1.0 W only if the percentage is not given.
(b) Change in temperature 15.7 and time 300 s used [M1]; 0.0523 K s^-1 [A1]. Allow 0.052. Allow °C s^-1.
(c) Use of c = P/(m × gradient) [M1]; 917 J kg^-1 K^-1 [A1]. Allow 918 if 0.0523 is used. ECF from (a)(i) and (b). Answer must be to 3 s.f. NOT 920.
(d) Percentages added: 1 + 2.1 + 3 [M1]; 6.1% [A1]; absolute uncertainty about 56 (accept 55 to 57, or 60 to 1 s.f.) and c = 920 ± 60 or 917 ± 56 [A1]. ECF from (a)(ii) and (c).
(e) Range of values 861 to 973 (or 897 compared with 917 ± 56) [B1]; states agrees / consistent as 897 lies within the range [B1]. ECF from (d). Allow a comment that a systematic error such as energy loss cannot be ruled out.
(f)* Level 3 (5 to 6 marks): the effect of energy loss on c is stated and explained correctly, a valid way of detecting losses from the data or graph is described, and two appropriate improvements are given. The reasoning is logical and connects the strands, and the explanation is supported by physics.
Level 2 (3 to 4 marks): two of the three strands are addressed with some supporting reasoning, or all three are addressed with limited explanation or one error.
Level 1 (1 to 2 marks): one strand or isolated relevant statements, little or no explanation.
Level 0 (0 marks): no creditworthy response.
Indicative points: energy is lost to the surroundings so less energy heats the block; the temperature rise is smaller; the gradient is smaller; the calculated c is too large (ORA); losses increase as the temperature difference to the surroundings increases; the graph would curve (gradient decreases) rather than be straight; compare the gradient at the start and end of the heating; compare the energy supplied VIt with mcΔθ; more or better insulation; a lid on the block; thermal paste or oil in the holes; start below room temperature so the final temperature is above it by a similar amount; repeat and average; use a data logger or more precise thermometer. Ignore: 'human error', 'use a more accurate meter' without detail.

\(R = 6.0/0.50 = 12\,\Omega\). Percentage uncertainty in \(V\) = \(0.2/6.0 \times 100 = 3.3\%\); in \(I\) = \(0.01/0.50 \times 100 = 2.0\%\). For a quotient the percentage uncertainties add: \(5.3\%\). Absolute uncertainty = \(0.053 \times 12 = 0.64 \approx 0.6\,\Omega\). Answer B.

B1: B selected. Method (for reference): M1 add percentage uncertainties (3.3% + 2.0%); A1 convert to absolute uncertainty 0.6 Ω. A (0.4) uses only the uncertainty in V; C doubles the result; D is the uncertainty in V as an absolute value. Only the letter B is credited.

\(F = BIL\sin\theta = 0.40 \times 3.0 \times 0.20 \times \sin 30^\circ = 0.24 \times 0.5 = 0.12\,\text{N}\). Answer C.

B1: C selected. Method: M1 use of \(F = BIL\sin\theta\) with the angle between wire and field; A1 0.12 N. A omits the angle; B uses cos 30°; D includes an extra factor of 0.5 (sin applied twice). Only the letter C is credited.

Energy supplied \(= Pt = 800 \times 105 = 8.4 \times 10^4\,\text{J}\). \(E = mc\Delta\theta\) so \(\Delta\theta = 8.4\times10^4 / (0.50 \times 4200) = 40\,\text{K}\). Answer D.

B1: D selected. Method: M1 \(E = Pt\) equated to \(mc\Delta\theta\); A1 40 K. Accept 40 °C for a temperature change. A, B and C arise from omitting or mishandling the mass or time.

\(I = I_0 e^{-\mu x}\), so \(I/I_0 = e^{-0.35 \times 4.0} = e^{-1.4} = 0.247 \approx 0.25\). Answer A.

B1: A selected. Method: M1 use of \(I = I_0e^{-\mu x}\) with \(\mu x = 1.4\); A1 0.25. C is \(e^{-0.35}\) (thickness ignored); D is \(e^{-0.15}\)-type error; B is a reciprocal-style error.

\(640 \to 320 \to 160 \to 80\) is three half-lives. So \(3t_{1/2} = 45\) minutes and \(t_{1/2} = 15\) minutes. Answer B.

B1: B selected. Method: M1 identify three half-lives (ratio 8); A1 15 minutes. A divides by 8; C treats the fall as a single halving; D ignores the decay.

\(\Delta\lambda/\lambda \approx v/c = 1.5\times10^{7} / 3.00\times10^{8} = 0.050\). \(\Delta\lambda = 0.050 \times 500 = 25\,\text{nm}\). Answer C.

B1: C selected. Method: M1 use of \(\Delta\lambda/\lambda \approx v/c\); A1 25 nm. A and D are powers-of-ten slips in \(v/c\) or the factor of 2.

\(x = \lambda D / a = (600\times10^{-9} \times 2.0) / (0.30\times10^{-3}) = 4.0\times10^{-3}\,\text{m} = 4.0\,\text{mm}\). Answer D.

B1: D selected. Method: M1 use of \(\lambda = ax/D\) rearranged for \(x\) with correct unit conversions; A1 4.0 mm. A, B and C come from incorrect powers of ten or halving the result.

\(V = -GM/r\), so at \(2R\) the potential is half as negative: \(-2.0\times10^{7}\,\text{J kg}^{-1}\). \(\Delta V = +2.0\times10^{7}\,\text{J kg}^{-1}\). \(\Delta E = m\Delta V = 25 \times 2.0\times10^{7} = 5.0\times10^{8}\,\text{J}\). Answer A.

B1: A selected. Method: M1 potential at \(2R\) is half the surface value; M1 \(\Delta E = m\Delta V\); A1 \(5.0\times10^{8}\) J. B uses a quarter of V; C uses the full surface value; D doubles it.

Percentage uncertainty in \(d\) = \(0.02/0.80 \times 100 = 2.5\%\). Since \(A \propto d^2\), the percentage uncertainty in \(A\) is doubled: \(2 \times 2.5\% = 5.0\%\). Answer C.

C (1 mark). B1 for correct answer only. Common errors: A (power not applied), D (percentage squared, 2.5² = 6.25).

Percentage uncertainty in \(L\) = \(0.005/0.500 = 1.0\%\). Percentage uncertainty in \(T\) = \(0.02/1.42 = 1.4\%\). Because \(T\) is squared, its contribution is \(2 \times 1.4 = 2.8\%\). Total = \(1.0 + 2.8 = 3.8\%\).

A (1 mark). B1 for correct answer only. B: power of T not applied (1.0 + 1.4). C: contribution of L omitted.

\(F = BIL\sin\theta = 0.40 \times 3.0 \times 0.15 \times \sin 30^\circ = 0.18 \times 0.5 = 0.090\,\text{N}\).

B (1 mark). B1 for correct answer only. C: cos 30° used. D: sin θ omitted. A: factor of 2 error.

\(I = I_0 e^{-\mu x}\) so \(0.25 = e^{-12.0\mu}\). Then \(\mu = \ln 4 / 12.0 = 1.386/12.0 = 0.12\,\text{mm}^{-1}\).

D (1 mark). B1 for correct answer only. A: ln 2 used (50% transmitted). B: 1/12 used.

\(640 \to 320 \to 160 \to 80 \to 40 \to 20\) is 5 half-lives. Half-life = \(30/5 = 6.0\) minutes.

C (1 mark). B1 for correct answer only. A: 4 half-lives assumed. B: 3 half-lives assumed. D: 7 half-lives assumed.

\(\lambda = \frac{ax}{D} = \frac{0.50 \times 10^{-3} \times 2.4 \times 10^{-3}}{2.0} = 6.0 \times 10^{-7}\,\text{m}\).

D (1 mark). B1 for correct answer only. A: extra factor of 2 in denominator. B: factor of 2 in numerator.

Vertical motion: \(20 = \tfrac{1}{2} g t^2\) so \(t = \sqrt{40/9.81} = 2.02\,\text{s}\). Horizontal distance = \(12 \times 2.02 = 24\,\text{m}\).

A (1 mark). B1 for correct answer only. D: time taken as h/v-type misuse. C: time doubled.

Place the heater and thermometer in their holes (a drop of oil in the thermometer hole improves thermal contact). Measure the mass \( m \) of the cylinder with the balance. Wrap the cylinder in insulation. Connect the heater to the supply with the ammeter in series and the voltmeter in parallel with the heater, and record the initial temperature. Switch on and start the stopwatch. Record the p.d. \( V \), current \( I \) and the temperature \( \theta \) at regular intervals (e.g. every 60 s) for about 10 minutes, keeping \( V \) and \( I \) constant. Allow time for the heat to spread through the block before reading the thermometer.

Energy supplied \( = VIt \) and \( VIt = mc\Delta\theta \), so \( \theta = \dfrac{VI}{mc}\,t + \theta_0 \). Plot \( \theta \) (y-axis) against \( t \) (x-axis). The gradient is \( \dfrac{VI}{mc} \), so \( c = \dfrac{VI}{m \times \text{gradient}} \).

Systematic error: thermal energy is lost to the surroundings, so the measured temperature rise is too small and \( c \) is overestimated. Insulation reduces the loss; also starting the block a few degrees below room temperature and ending the same amount above it makes gains and losses roughly cancel. Using the gradient of the graph also reduces the effect of the thermometer not being in equilibrium initially.

Extended response, level of response, 6 marks. Indicative scientific points (not a compulsory script):

1. Mass of cylinder measured with the balance (B).
2. Heater and thermometer in holes, with good thermal contact (e.g. oil) and insulation around the cylinder (B).
3. Measure V and I (or joule meter reading) and the time with a stopwatch, with temperature recorded at regular intervals while the power is constant (B).
4. Use of \( VIt = mc\Delta\theta \) (energy supplied = thermal energy gained) (B).
5. Plot \( \theta \) against \( t \) with gradient \( = VI/(mc) \), so \( c = VI/(m \times \text{gradient}) \) (B). Allow plot of energy against \( \theta \) with gradient \( mc \).
6. Systematic error identified (energy loss to surroundings, or heat capacity of heater/thermometer, or thermal lag) and a valid reduction (insulation, starting below and finishing above room temperature, or allowing time for conduction through block before reading) (B).

Band 3 (5 to 6 marks): the procedure is complete, logical and sequenced, measurements and the method of analysis are both correct and linked by the correct equation, and a systematic error with its reduction is clearly given.
Band 2 (3 to 4 marks): a mostly correct procedure with measurements and some analysis, but one strand is missing, unclear or not linked (e.g. no graph or no gradient link, or no systematic error).
Band 1 (1 to 2 marks): a few relevant points such as measuring mass, p.d., current, time or temperature without a coherent method.
0 marks: no creditworthy response.

Allow: use of a joule meter for energy. Ignore: repeat readings and averaging unless linked to a systematic error. NOT: a reduction in 'random' error given as the systematic-error reduction. AW for all points.

(a) \(A = \pi d^2/4 = \pi (0.320\times10^{-3})^2/4 = 8.04\times10^{-8}\ \text{m}^2\). Percentage uncertainty in \(d\) = \(0.010/0.320 = 3.1\%\). \(A \propto d^2\), so the percentage uncertainty in \(A\) is \(2\times3.125 = 6.25\% \approx 6.3\%\).

(b) \(R = V/I = 1.45/0.250 = 5.80\ \Omega\). Percentage uncertainty in \(V\) = 0.69%; in \(I\) = 2.0%. Total = 2.7%.

(c) \(\rho = (5.80\times8.04\times10^{-8})/0.850 = 5.49\times10^{-7}\ \Omega\,\text{m}\). Percentage uncertainty in \(L\) = 0.24%. Total = \(6.25+2.7+0.24 = 9.2\%\). Absolute uncertainty = \(0.092\times5.49\times10^{-7} = 0.5\times10^{-7}\). So \(\rho = (5.5\pm0.5)\times10^{-7}\ \Omega\,\text{m}\).

(d) The diameter contributes most (6.3% of the 9.2% total) because it is squared. Take readings at several positions along the wire and in perpendicular directions with a micrometer, and use the mean. This reduces the effect of non-uniformity and random error.

(a) B1: \(A = 8.04\times10^{-8}\ \text{m}^2\) (allow 8.0 and \(0.0804\ \text{mm}^2\) if unit consistent). B1: 6.3% or 6.25% (doubling the percentage uncertainty in \(d\)). NOT 3.1% only.

(b) B1 (M1): \(R = 5.80\ \Omega\). A1: 2.7% (accept 2.69%; from 0.69% + 2.0%). Allow 3% (1 s.f.) only if working shown.

(c) M1: \(\rho = RA/L\) with correct substitution. A1: \(\rho = 5.5\times10^{-7}\ \Omega\,\text{m}\) (allow 5.49; ECF from (a) and (b)). A1: uncertainty \(\pm0.5\times10^{-7}\ \Omega\,\text{m}\) (from summed percentages, about 9.2%, ECF), with the uncertainty given to 1 s.f. and \(\rho\) to the same decimal place. Ignore uncertainty given to 2 s.f. if the value is consistent, but do not award the final A1 if the decimal places do not match.

(d) B1: the diameter (AW: \(d\) or area). Allow ECF from (a)-(c) if the percentage uncertainties support the choice. B1: a valid improvement, for example repeat readings of \(d\) at different points along the wire and in different orientations, and average; or use a thicker wire; or use a micrometer with finer resolution. NOT 'use more accurate apparatus' without detail. Ignore changes to \(L\), \(V\) or \(I\).

(a) \(A = \pi d^2/4\), so percentage uncertainty in \(A\) is \(2\times(0.01/0.38)\times100 = 2\times2.63 = 5.3\%\).

(b) \(A = \pi(0.38\times10^{-3})^2/4 = 1.13\times10^{-7}\ \mathrm{m^2}\). \(\rho = RA/L = 5.2\times1.134\times10^{-7}/0.800 = 7.4\times10^{-7}\ \Omega\) m.

(c) Percentage uncertainty in \(\rho\) = \(3.85\% + 5.3\% + 0.25\% = 9.4\%\). Absolute uncertainty = \(0.094\times7.37\times10^{-7} = 0.7\times10^{-7}\ \Omega\) m.

(d) Measure d at several points along and around the wire with a micrometer, zero check first, and average. Measure L with a metre rule from contact to contact with the wire straight and taut. Vary current with a variable resistor, record V and I for several values, keep current small / switch off between readings to avoid heating (which changes resistance). Plot V against I, draw a line of best fit, and take the gradient. Graph averages random error, reveals anomalies and systematic offsets (zero error), whereas a single pair gives no check on reliability.

(a) B1: 5.3% (allow 5%; accept 5.26).

(b) M1: \(A = \pi d^2/4\) evaluated (\(1.1\times10^{-7}\) m\(^2\)) and use of \(\rho = RA/L\). A1: \(7.4\times10^{-7}\ \Omega\) m. Allow 7.3 to 7.4 if A rounded. NOT 1 s.f. answer for the A1.

(c) B1: \(\pm0.7\times10^{-7}\ \Omega\) m (percentage uncertainties 3.8 + 5.3 + 0.25 = 9.4%). ECF from (a) and (b).

(d)* Level of response, 6 marks.
Band 3 (5 to 6 marks): Practical method for at least two of d, L and R is clear and includes specific precautions; explanation of the advantage of the graph is complete and linked to reducing random error and identifying anomalies or systematic error. Reasoning is logical and coherent.
Band 2 (3 to 4 marks): Some correct practical detail for d, L or R and a partial explanation of the graph advantage; some gaps in logic.
Band 1 (1 to 2 marks): Isolated relevant statements with little linkage.
0 marks: no creditworthy response.
Indicative points: repeat d readings at different positions and orientations and average; check micrometer zero / correct zero error; ratchet used; L measured between contact points with wire straight; ruler with suitable resolution; vary the current using a variable resistor / potential divider and record several (V, I) pairs; keep current low or switch off between readings to avoid heating; line of best fit through data; gradient averages out random errors; anomalous points are identified; straight line through origin confirms ohmic behaviour; a non-zero intercept reveals systematic error; uncertainty in gradient can be found from worst-fit lines. Ignore references to repeating readings without averaging.

(a) At the half-value thickness \(I = I_0/2\), so \(\tfrac12 = e^{-\mu x_{1/2}}\), giving \(\mu = \ln 2/x_{1/2} = 0.693/2.4 = 0.289 \approx 0.29\ \mathrm{mm^{-1}}\).

(b) \(I/I_0 = e^{-0.2888\times7.0} = e^{-2.02} = 0.13\).

(c) X-rays from a source pass through the arm and are detected by a film/digital detector behind it. Bone has a larger attenuation coefficient (higher density and effective atomic number, so more photoelectric absorption) than soft tissue, so less intensity is transmitted through bone, leaving a lighter/less exposed region. The difference in transmitted intensity gives contrast, and a fracture shows as a break in the bone shadow. The aluminium filter absorbs low-energy X-ray photons, which are strongly absorbed by the body and contribute to dose but not to the image. This reduces patient dose and gives a harder beam with better image quality.

(a) M1: \(\tfrac12 = e^{-\mu x}\) with \(x = 2.4\) mm, or \(\ln 2 = \mu x\). A1: \(\mu = 0.693/2.4 = 0.289\) (shown to at least 2 s.f. before rounding to 0.29). Allow use of any numerical check that reproduces 0.29.

(b) C1: \(\mu x = 0.289\times7.0\) (= 2.02), or alternatively 7.0/2.4 = 2.9 half-lives. A1: 0.13 or 13%. Allow 0.12 to 0.14 (ignore rounding of \(\mu\) to 0.29 which gives 0.131). NOT a value of intensity in W m\(^{-2}\).

(c)* Level of response, 6 marks.
Band 3 (5 to 6 marks): Describes image formation with detector and shadow; explains contrast through different attenuation of bone and soft tissue with a reason; explains the filter removing low-energy photons and reducing dose. Reasoning is clear, linked and logically ordered.
Band 2 (3 to 4 marks): Two of the three strands addressed with some correct physics, or all three with partial explanations.
Band 1 (1 to 2 marks): Isolated correct statements, for example bone absorbs more X-rays.
0 marks: no creditworthy response.
Indicative points: X-rays produced and directed at the arm; detector or film placed on the opposite side; image is a shadow of transmitted intensity; bone has larger \(\mu\) (greater density / higher proton number, more photoelectric absorption); bone transmits lower intensity so appears lighter on the image (AW depending on detector type); fracture appears as a gap or line in the bone image; intensity falls exponentially with thickness; low-energy photons are absorbed more readily by tissue; filter removes these, which would add to dose without contributing to the image; reduces patient dose; narrower range of photon energies improves contrast / image quality. Allow ORA for soft tissue. Ignore references to ionisation damage unless linked to dose.

(a) \(E = Pt = 60\times300 = 1.8\times10^4\) J.

(b) \(E = mc\Delta\theta\), so \(c = 18000/(1.20\times16.0) = 938\ \mathrm{J\,kg^{-1}\,K^{-1}}\), about 940.

(c) Some energy is transferred to the surroundings and to the heater and thermometer, so the energy actually gained by the block is less than 18 000 J. The temperature rise is therefore smaller than expected if all energy went to the block, so \(c = E/(m\Delta\theta)\) with \(E\) too large gives a value that is too large.

(d) Wrap the block in thermal insulation (lagging).

(e)(i) \(E_k = \tfrac32 kT = 1.5\times1.38\times10^{-23}\times300 = 6.21\times10^{-21}\) J.

(ii) \(\tfrac12 m\langle c^2\rangle = 6.21\times10^{-21}\), so \(\langle c^2\rangle = 1.87\times10^6\ \mathrm{m^2\,s^{-2}}\) and \(c_{rms} = 1.37\times10^3\ \mathrm{m\,s^{-1}}\).

(a) B1: \(1.8\times10^4\) J (18 000 J). Unit required.

(b) C1: \(c = E/(m\Delta\theta)\) with \(\Delta\theta = 16.0\) K. A1: 938 (J kg\(^{-1}\) K\(^{-1}\)); allow 940 or 9.4 x 10^2. ECF from (a). Allow J kg\(^{-1}\) \(^\circ\)C\(^{-1}\).

(c) B1: energy is lost (transferred) to the surroundings / not all the energy supplied heats the block (AW). B1: the energy used in the calculation is larger than the energy that actually raised the temperature of the block, so the calculated \(c\) is too large (AW; ORA). NOT 'heat is lost' alone for the second mark.

(d) B1: insulate / lag the block (accept lid, or add oil in the holes for better thermal contact). Ignore 'repeat readings'.

(e)(i) C1: \(E = \tfrac32 kT\). A1: \(6.21\times10^{-21}\) J. Allow \(6.2\times10^{-21}\) J.

(e)(ii) C1: \(\tfrac12 m\langle c^2\rangle = E\) (or \(c_{rms} = \sqrt{3kT/m}\)). A1: \(1.37\times10^3\ \mathrm{m\,s^{-1}}\) (allow 1.4 x 10^3). ECF from (e)(i).

(a)(i) \(\ln I = \ln I_0 - \mu x\), so plot \(\ln I\) (y-axis) against \(x\) (x-axis); gradient \(= -\mu\).

(ii) \(\ln 143 = 4.963\).

(iii) Gradient \(= (4.466 - 6.461)/(8.0 - 0) = -0.249\ \text{mm}^{-1}\), so \(\mu = 0.249 \approx 0.25\ \text{mm}^{-1}\).

(b) \(0.050 = e^{-\mu x}\), so \(x = \ln(20)/\mu = 2.996/0.249 = 12.0\) mm (12 mm using \(\mu = 0.25\)).

(c) X-rays pass through the arm and are attenuated according to \(I = I_0 e^{-\mu x}\). Bone has a much larger attenuation coefficient (higher density and higher proton number, so more photoelectric absorption) than soft tissue, so less intensity is transmitted through bone and the detector or film shows bone as a lighter region (shadow). Different soft tissues have very similar densities and attenuation coefficients, so the transmitted intensities are almost equal and contrast is poor. A contrast medium such as a barium meal (high atomic number, strongly absorbing) fills the organ, e.g. the digestive tract, making it show clearly. Low-energy X-rays are absorbed by the body without contributing to the image and increase the dose; an aluminium filter preferentially absorbs these, so the patient dose is reduced and the beam is more nearly monochromatic and penetrating.

(a)(i) B1: \(\ln I\) on the y-axis against \(x\) on the x-axis. Allow \(\lg I\) against \(x\). NOT \(I\) against \(x\).

(a)(ii) B1: 4.963. Allow 4.96.

(a)(iii) M1: gradient \(= (4.466 - 6.461)/8.0\) (or any correct use of two points), with \(\mu = -\)gradient. A1: \(\mu = 0.25\ \text{mm}^{-1}\) (accept 0.249 and range 0.24 to 0.26). Allow ECF for a wrong-sign gradient with M1 only.

(b) C1: \(\ln(0.050) = -\mu x\) or \(x = \ln 20/\mu\). A1: 12 mm (accept 11.9 to 12.1). ECF from (a)(iii) for \(\mu\). NOT use of 0.95 in place of 0.050.

(c)* Level of response, 6 marks.

5 to 6 marks: A clear, logically structured explanation covering all three bullets, with correct supporting physics (attenuation coefficient or density/proton number differences; high-Z contrast medium; filter removing low-energy photons so that dose is reduced). The strands are linked, e.g. contrast arises from differing transmitted intensities. Few or no errors.

3 to 4 marks: Explanation covers at least two bullets with some correct physics, or all three with limited supporting detail or a minor error. Reasoning is mostly coherent.

1 to 2 marks: Isolated correct statements, e.g. bone absorbs more X-rays than tissue, or barium improves contrast, with little explanation or linking.

0 marks: No creditworthy response.

Indicative scientific points: X-rays are attenuated exponentially, \(I = I_0 e^{-\mu x}\); bone has a larger \(\mu\) (greater density and atomic number; photoelectric absorption increases with \(Z\)); less transmission through bone gives a lighter (unexposed) region on film or a lower detector signal; soft tissues have similar \(\mu\), so similar transmitted intensities and low contrast; contrast medium such as barium or iodine compounds, strongly absorbing, high \(Z\); the filter absorbs low-energy (soft) X-rays which would be absorbed by the patient and not reach the detector, so it reduces dose and hardens the beam. Allow: use of a collimator or anti-scatter grid as an extra creditworthy point for sharpness. Ignore: references to ionisation risk without link to dose. NOT: X-rays 'reflected' by bone.

(a)(i) Corrected count rate at \(t=0\): \(128-8 = 120\ \text{s}^{-1}\). At 50 min: \(38-8 = 30\ \text{s}^{-1}\). \(30/120 = 1/4 = (1/2)^2\), so 50 min is two half-lives, giving \(T_{1/2} = 50/2 = 25\) min.

(ii) \(T_{1/2} = 25\times60 = 1500\ \text{s}\). \(\lambda = \ln 2 / T_{1/2} = 0.693/1500 = 4.62\times10^{-4}\ \text{s}^{-1}\).

(iii) \(A = \lambda N\) so \(N = 120 / (4.62\times10^{-4}) = 2.6\times10^{5}\) nuclei.

(b) Indicative method: measure the background count rate with no source and subtract it from all readings. Place the GM tube close to the source and record the corrected count rate. Place paper between source and tube: a large fall indicates alpha. Replace with aluminium sheets (a few millimetres): a further fall indicates beta. Then place thick lead: any remaining drop indicates gamma, since gamma is only gradually attenuated; a count rate unchanged by lead means little or no gamma. Repeat readings over a fixed time and average to reduce random error. Alpha has a range of only a few centimetres in air, so a count rate that falls to background at a distance of about 5 cm or more suggests alpha only; the inverse-square variation of the count rate with distance can distinguish gamma. Safety: handle the source with tongs, never by hand; keep it at the maximum practical distance and in its shielded container when not in use; minimise the exposure time; point the source away from people; wear gloves and wash hands afterwards; do not eat or drink in the area; keep a record of the source and store it in a locked lead-lined box.

(a)(i) B1: subtracts background from both readings (120 and 30 s^-1). M1: identifies the ratio 30/120 = 1/4 as two half-lives (or equivalent halving method: 120 -> 60 -> 30). A1: T1/2 = 25 min (shown clearly, 50/2). Allow ECF only if background is subtracted. NOT 128 and 38 used directly (gives a different value).

(ii) C1/M1: converts 25 min to 1500 s AND uses \(\lambda = \ln 2 / T_{1/2}\). A1: \(4.6\times10^{-4}\ \text{s}^{-1}\) (accept \(4.62\times10^{-4}\)). Allow 0.028 min^-1 for M1 only if unit conversion missing (A0).

(iii) M1: \(N = A/\lambda\) with A = 120 s^-1. A1: \(2.6\times10^{5}\) (accept 2.59-2.60 x 10^5). ECF from (ii). NOT use of 128 s^-1 (gives 2.8 x 10^5, M1 only).

(b)* Level of response, six marks, assessed holistically.

Level 3 (5-6 marks): A logical, complete method is described covering background correction, a sequence of absorbers (paper, aluminium, lead) with what the changes in count rate show for each radiation type, AND at least two distinct, correctly explained safety measures (e.g. tongs, distance, time, shielded storage). Reasoning is clear and connects the observations to the conclusions about the radiation types.

Level 2 (3-4 marks): Method describes use of at least two types of absorber with a mostly correct interpretation, or gives a good method with limited safety, or good safety with a partial method. Some links between observation and conclusion are missing or lack clarity.

Level 1 (1-2 marks): Isolated relevant points, e.g. measures count rate with and without an absorber, or states one safety precaution, with little or no connection between them.

Level 0 (0 marks): No creditworthy response.

Indicative scientific points (not a compulsory script): measure and subtract background; record count rate with no absorber at fixed distance and time; paper stops alpha (large drop); a few mm of aluminium stops beta; thick lead needed to reduce gamma substantially; remaining count rate after lead close to background means no gamma; repeat and average readings; alpha range a few cm in air; use tongs, do not hold or point the source at anyone; store in a shielded container when not in use; keep distance and minimise time; wash hands; record keeping.

Allow: AW for all terms. Ignore: references to 'radiation sickness' without explanation. NOT: 'lead stops all radiation'; claims that the GM tube can distinguish energies.

(a) Mean = (0.38 + 0.37 + 0.39 + 0.38 + 0.38)/5 = 0.38 mm. Range = 0.39 − 0.37 = 0.02 mm, so uncertainty = 0.01 mm. So d = 0.38 ± 0.01 mm.

(b)(i) \(A = \pi d^2/4 = \pi (0.38\times10^{-3})^2/4 = 1.13\times10^{-7}\ \text{m}^2\).

(b)(ii) % uncertainty in d = (0.01/0.38) × 100 = 2.6%. Since \(A \propto d^2\), % uncertainty in A = 2 × 2.6 = 5.3%.

(c)(i) \(E = \dfrac{\text{stress}}{\text{strain}} = \dfrac{F/A}{x/L} = \dfrac{F}{x}\cdot\dfrac{L}{A}\), so \(E = \text{gradient} \times L/A\). \(E = 9.1\times10^{3} \times 2.500 / 1.13\times10^{-7} = 2.0\times10^{11}\) Pa.

(c)(ii) % uncertainty in gradient = (9.1 − 8.7)/9.1 × 100 = 4.4%. % uncertainty in L = 0.005/2.500 × 100 = 0.2%. Total = 4.4 + 0.2 + 5.3 = 9.9% (≈ 10%), i.e. E = (2.0 ± 0.2) × 10^11 Pa.

(d) The diameter contributes the most (5.3%) because A depends on d², so its percentage uncertainty is doubled and the micrometer reading is only about 0.38 mm, so the 0.01 mm uncertainty is relatively large. The gradient contributes 4.4%, which comes from the scatter of the points about the line. The length contributes only 0.2% and is negligible. Improvements: take more diameter readings at different positions and in perpendicular orientations and use the ratchet; check and correct for zero error; use a longer wire or a more sensitive method (travelling microscope, vernier scale or reference wire) to measure larger extensions with smaller percentage uncertainty; take more readings over a wider range of load and repeat; use a graph with a line of best fit; do not exceed the elastic limit, check by unloading and confirming that the extension returns to zero; apply an initial small load to remove kinks.

(a) B1: mean d = 0.38 mm. B1: uncertainty = 0.01 mm (half of range 0.02 mm). Allow 0.38 ± 0.01 mm as one statement for both marks. [2]

(b)(i) B1: A = 1.1 × 10^-7 m^2 (allow 1.13 × 10^-7). Allow ECF for d from (a). NOT if mm^2 used without conversion. [1]

(b)(ii) M1: % uncertainty in d = 0.01/0.38 × 100 = 2.6% (allow 2.6–2.7%). A1: doubled for d², so 5.3% (allow 5.2–5.4%; ECF from (a)). Using the 'absolute uncertainty in A' route is acceptable if correct. [2]

(c)(i) M1: E = (F/A)/(x/L) rewritten as gradient × L/A, or substitution of 9.1 × 10^3, 2.500 and the area. A1: E = 2.0 × 10^11 Pa (must show an unrounded value, e.g. 2.01 × 10^11, or at least 3 s.f., to be credited). Ignore unit Pa = N m^-2. [2]

(c)(ii) M1: percentage uncertainties of gradient (4.4%, allow 4.3–4.4%) and L (0.2%) found and added to the % uncertainty in A. A1: 9.9% (allow 10%); ECF from (b)(ii). Allow absolute uncertainty ±0.2 × 10^11 Pa as an alternative to the percentage if clearly the result of the correct addition. NOT combining by subtraction. [2]

(d)* Level of response, assessed holistically. [6]

Indicative scientific points:
- % uncertainty in d (2.6%) is doubled to 5.3% in A as A ∝ d², so d is the largest contributor.
- Gradient uncertainty (4.4%) is the second largest; due to scatter / limited range of readings.
- Uncertainty in L (0.2%) is negligible.
- Diameter: more readings at different positions / orientations (perpendicular directions); use the ratchet; check zero error; average the readings. Allow a thicker wire, but only if the extension is still measurable.
- Extension: use a longer wire, larger loads (within the elastic limit) or a travelling microscope / vernier scale to measure the extension with a smaller percentage uncertainty.
- Gradient: more data points over a larger range, repeat readings for each load, use a best-fit line.
- Elastic behaviour: do not exceed the elastic limit; unload and check that the extension returns to zero; the graph should be linear through the origin; use an initial load to remove kinks.

5–6 marks: A sound evaluation that ranks all three contributions with quantitative support (d largest because squared, then gradient, L negligible) and gives at least three relevant, justified improvements, including one linked to the elastic limit. The reasoning is logically structured and clearly links each improvement to a reduction in uncertainty.
3–4 marks: Identifies the main contributions (at least d and gradient) with some quantitative or reasoned support, and gives at least two relevant improvements. Some reasoning may be incomplete, or the elastic behaviour may be omitted.
1–2 marks: Makes limited relevant statements, such as one contribution or one or two improvements, with little or no justification or linking.
0 marks: No creditworthy response.

Allow ECF from earlier values in the evaluation of contributions. Ignore references to human error or to 'using a more accurate micrometer' without justification.

(a)(i) Energy supplied \(E = VIt = 12.0 \times 3.00 \times 300 = 1.08\times10^{4}\) J. \(c = E/(m\Delta\theta) = 10800/(0.500\times 24.0) = 900\ \mathrm{J\,kg^{-1}\,K^{-1}}\).

(ii) Percentage uncertainties: \(V\): \(0.1/12.0 = 0.83\%\); \(I\): \(0.05/3.00 = 1.67\%\); \(t\): \(1/300 = 0.33\%\); \(m\): \(0.002/0.500 = 0.40\%\); \(\Delta\theta\): \(1.0/24.0 = 4.17\%\). Total \(= 7.4\%\). Absolute uncertainty \(= 0.074\times 900 = 67\), so \(\approx 70\ \mathrm{J\,kg^{-1}\,K^{-1}}\) (1 s.f.), i.e. \(c = (9.0\pm0.7)\times10^{2}\).

(b) The temperature rise has by far the largest percentage uncertainty (4.2%). Heat for longer (or use a thermometer with finer resolution) so that \(\Delta\theta\) is larger and the 1.0 K uncertainty is a smaller fraction of it.

(c) Some of the energy supplied is lost to the surroundings, so the energy gained by the block is less than \(VIt\). The temperature rise is smaller than it should be, so the calculated \(c\) is too large.

(a)(i) M1: \(E = VIt\) (= 10 800 J) used in \(c = E/(m\Delta\theta)\). A1: 900 \(\mathrm{J\,kg^{-1}\,K^{-1}}\). Allow \(9.0\times10^{2}\). Ignore number of significant figures of 2 or more.

(a)(ii) M1: calculates the fractional or percentage uncertainty in each of the five quantities (at least four correct). M1: adds the percentage uncertainties (7.4%; allow 7.3 to 7.5%). A1: absolute uncertainty = 67 (allow 66 to 68), quoted to 1 or 2 s.f. (70 or 67). NOT 3 or more s.f. NOT adding absolute uncertainties directly. ECF allowed for the final A1 from the candidate's total percentage uncertainty.

(b) B1: heat for longer / produce a larger temperature rise / use a thermometer with smaller resolution AND reason that the percentage uncertainty in \(\Delta\theta\) is the largest or would be reduced. AW. Ignore improvements to \(m\), \(V\), \(I\) or \(t\) only. Both change and reason are required.

(c) B1: calculated \(c\) is too large (greater than true value) because the energy supplied is greater than that gained by the block / temperature rise is smaller than expected. ORA. NOT 'inaccurate' without direction.

(a) Use \(I = I_0 e^{-\mu x}\). Then \(0.35 = e^{-\mu \times 4.0}\), so \(\mu = \ln(1/0.35)/4.0 = 1.050/4.0 = 0.262\,\text{mm}^{-1}\) (0.26 mm^-1 to 2 s.f.).

(b) Bone has a much larger attenuation coefficient than soft tissue because it is denser and contains atoms of higher proton number (calcium), so photoelectric absorption is greater. Intensity transmitted through bone is therefore much lower than through the surrounding tissue, giving contrast between bone and soft tissue on the detector. The aluminium sheet preferentially absorbs low-energy X-ray photons, which have a larger attenuation coefficient and would be absorbed in the patient's tissue, increasing the dose without reaching the detector. The beam passing into the patient is then of higher mean energy, reducing patient dose and improving image sharpness. Soft tissues in the digestive system have almost the same attenuation coefficient as neighbouring tissue so produce very little contrast. Barium (high proton number) has a very high attenuation coefficient, absorbs far more X-rays than the surrounding tissue, and so the barium-filled organ shows clearly against the background; the barium is non-toxic in an insoluble form and is not absorbed into the body.

(a) 3 marks
M1: use of \(I = I_0 e^{-\mu x}\) with \(I/I_0 = 0.35\) and \(x = 4.0\) mm (or \(\mu x = \ln(1/0.35)\)). [1]
M1: correct rearrangement/use of natural logarithm, \(\ln(1/0.35)\) or \(-\ln 0.35 = 1.05\). [1]
A1: \(\mu = 0.26\,\text{mm}^{-1}\) (0.262 mm^-1). Allow 0.25 to 0.27 only if working shown. NOT a negative value. Allow 262 m^-1 (or 260 m^-1) if the unit is stated. NOT 0.026 mm^-1. [1]

(b)* 6 marks. Level of response marking, assessed holistically.

Indicative scientific points:
- Bone has a larger attenuation (absorption) coefficient than soft tissue / bone absorbs more X-rays (higher density and/or higher proton number, calcium).
- Lower intensity transmitted through bone than through soft tissue, giving contrast (bone appears lighter/white on the film or detector: AW).
- Aluminium sheet removes/absorbs low-energy (soft) X-ray photons.
- Low-energy photons would be absorbed by the patient's body and not reach the detector, so they increase dose without improving the image.
- Beam is hardened / mean photon energy raised, so dose is reduced and image quality is maintained.
- Soft tissue of the digestive system has similar attenuation to neighbouring tissue so there is little contrast.
- Barium has a high proton number / high attenuation coefficient, so it absorbs strongly and outlines the organ (barium meal is a contrast medium).
- Barium is chemically inert/insoluble so safe to swallow (allow, not essential).

Band 3 (5-6 marks): All three requirements addressed with correct, clearly linked physics; explanations connect attenuation coefficient or absorption to the observed contrast or dose reduction. Reasoning is coherent and logically structured with few or no omissions.
Band 2 (3-4 marks): At least two requirements addressed with some correct physics, or all three addressed but with limited explanation (for example statements without a link to attenuation or absorption). Reasoning is mostly logical but may have gaps.
Band 1 (1-2 marks): One requirement addressed, or isolated correct statements with little linkage or explanation.
Band 0 (0 marks): No response or no creditworthy physics.

Guidance: Allow 'denser' and 'higher atomic number' for bone. Ignore references to X-ray wavelength being 'absorbed by' ionisation unless contradicting. NOT 'bone reflects X-rays'. NOT 'aluminium sheet stops all X-rays' or 'blocks the harmful rays' without reference to low-energy photons. NOT 'barium emits X-rays'.

(a) \(P = VI = 12.0 \times 3.50 = 42.0\) W. \(E = Pt = 42.0 \times 300 = 1.26\times10^{4}\) J. \(c = E/(m\Delta\theta) = 1.26\times10^{4}/(1.00\times14.0) = 900\ \mathrm{J\,kg^{-1}\,K^{-1}}\).

(b) Percentage uncertainties: \(V\): \(0.1/12.0 = 0.83\%\); \(I\): \(0.05/3.50 = 1.43\%\); \(t\): \(1/300 = 0.33\%\); \(m\): \(0.01/1.00 = 1.0\%\); \(\Delta\theta\): \(1.0/14.0 = 7.14\%\). Sum = 10.7%. Absolute uncertainty = \(0.107 \times 900 \approx 96\), so \(c = (900 \pm 100)\ \mathrm{J\,kg^{-1}\,K^{-1}}\).

(c) The temperature rise has the largest percentage uncertainty (7%). Heating for longer (for example 600 s) doubles \(\Delta\theta\) so its percentage uncertainty halves; a thermometer with finer resolution would also help. Plot \(\theta\) against \(t\) at regular intervals and draw a line of best fit. The gradient is \(\Delta\theta/\Delta t\) and \(c = P/(m \times \text{gradient})\), which averages out random errors and allows anomalies to be identified. Because the block is not insulated, some energy is transferred to the surroundings, so the energy absorbed by the block is less than \(VIt\). The temperature rise is therefore smaller than it should be and the calculated \(c\) is too large. Lagging the block with insulating material (and a lid), and using a little oil in the thermometer hole for good thermal contact, reduces the error.

(a)
M1: \(P = VI = 42.0\) W (or \(E = VIt\)).
M1: \(E = 1.26\times10^{4}\) J and use of \(E = mc\Delta\theta\) rearranged for \(c\).
A1: \(c = 900\ \mathrm{J\,kg^{-1}\,K^{-1}}\). Allow 1 or 2 s.f. (9 × 10^2). Unit required for A1; NOT missing or wrong unit.
[3]

(b)
B1: percentage uncertainty in \(\Delta\theta\) = 7.1% (7.14%) AND at least three other correct percentage uncertainties seen (0.83%, 1.43%, 1.0%, 0.33%).
M1: percentages added (not subtracted, not combined in quadrature) to give about 10.7% (11%). Allow ECF from B1.
A1: absolute uncertainty about 96, quoted as \(c = (900 \pm 100)\) with unit; allow \(900 \pm 96\). Ignore \(\pm 90\) to \(\pm 100\) from rounding. ECF from M1.
[3]

(c)* Level of response, holistic.

Band 3 (5–6 marks): All three strands addressed with correct, supported physics. The temperature rise is identified as the dominant uncertainty with a change that clearly reduces its percentage effect; the graph method is described with the gradient and a correct expression for \(c\); the direction of the systematic error is correct and explained in terms of energy losses, with a suitable modification. The reasoning is logical and connected.

Band 2 (3–4 marks): Two strands addressed well, or all three addressed with some omissions or lack of explanation (for example no link between losses and the direction of error, or the gradient not related to \(c\)).

Band 1 (1–2 marks): Some relevant points made, usually from one strand only, with limited explanation or reasoning.

Band 0 (0 marks): No creditworthy response.

Indicative scientific points:
- \(\Delta\theta\) has the largest percentage uncertainty (about 7%)
- heat for longer / increase \(\Delta\theta\) so percentage uncertainty falls; or use thermometer of better resolution (e.g. 0.1 K)
- take temperature readings at regular time intervals and plot \(\theta\) against \(t\)
- gradient = \(\Delta\theta/\Delta t\); \(c = P/(m \times \text{gradient})\) (AW, e.g. \(c = VI/(m \times\) gradient))
- best-fit line reduces random error; anomalies identifiable
- energy is lost to the surroundings, so not all electrical energy raises the temperature of the block
- measured \(\Delta\theta\) too small, so \(c\) too large (ORA); systematic error
- lag the block / use insulation / lid; oil in thermometer hole for thermal contact
Allow: repeating the experiment and averaging as an additional point only if linked to reducing random error. Ignore: reference to human reaction time unless linked to \(t\). NOT: 'c is too small' without a consistent explanation; NOT: 'use a more accurate ammeter' as the main source of uncertainty.
[6]

Total [12]

(a) Take natural logs of \(I = I_0 e^{-\mu x}\): \(\ln I = \ln I_0 - \mu x\). This has the form \(y = c + mx\), so the graph is a straight line with intercept \(\ln I_0\) and gradient \(-\mu\) (the magnitude of the gradient is the attenuation coefficient).

(b) \(\mu = \dfrac{\ln(I_0/I)}{x} = \dfrac{\ln(100/30)}{8.0} = \dfrac{1.204}{8.0} = 0.150\ \text{mm}^{-1}\).

(c) \(x_{1/2} = \dfrac{\ln 2}{\mu} = \dfrac{0.693}{0.150} = 4.6\) mm.

(d) Indicative content: X-rays are attenuated differently by different tissues, so the transmitted intensity varies across the detector and forms a shadow image. Attenuation depends on the density and proton number of the material (\(I = I_0 e^{-\mu x}\)). Soft tissues have similar \(\mu\), so contrast between a hollow organ and its surroundings is poor. Barium has a large proton number, so it absorbs strongly; it fills the organ and gives high contrast. Low-energy photons are absorbed in the body and add to patient dose without reaching the detector. The aluminium filter preferentially absorbs these low-energy photons, so the beam is harder, the patient dose falls, and the image is less blurred and clearer.

(a) B1: takes logarithms of both sides / \(\ln I = \ln I_0 - \mu x\). B1: compares with \(y = mx + c\) and states gradient \(= -\mu\) (allow magnitude of gradient \(= \mu\)). Total 2.

(b) M1: \(\ln(I_0/I)\) or \(\ln(100/30)\) (or \(\ln 30 - \ln 100\)) used with \(x = 8.0\). A1: \(1.204/8.0\) evaluated. A1: \(0.15\ \text{mm}^{-1}\) (allow 0.150; accept 0.15 to 2 or 3 s.f.). A bare correct answer scores 3. Total 3. Ignore sign if magnitude is correct; NOT a negative value unless stated as the gradient.

(c) C1/M1: \(x_{1/2} = \ln 2/\mu\) or \(0.5 = e^{-\mu x}\). A1: 4.6 mm (4.62 using 0.150). ECF from (b). Total 2.

(d) Level of response, 6 marks. Indicative scientific points: (1) differential attenuation by tissues / bone absorbs more, so shadow image; (2) attenuation depends on density and proton number; (3) soft tissues have similar attenuation so poor contrast; (4) barium has high proton number so absorbs strongly (contrast medium) and outlines organ; (5) low-energy X-rays are absorbed by the body without contributing to the image; (6) filter removes low-energy photons, reducing patient dose; (7) filter gives a harder, more intense/narrower-energy beam so clearer image / sharper.
Band 3 (5-6 marks): explains image formation AND the role of both barium and filter with correct, linked physics (attenuation, proton number, photon energy, dose). Reasoning is logical, complete and well supported.
Band 2 (3-4 marks): addresses image formation and at least one of the two measures with some correct explanation, or all three at a descriptive level with limited linking.
Band 1 (1-2 marks): makes isolated relevant statements, e.g. barium absorbs X-rays or the filter removes some X-rays, with little explanation.
Band 0 (0 marks): no creditworthy response.
Allow AW throughout. Ignore references to ionising effects of X-rays unrelated to dose. NOT 'barium is radioactive'.

(a)(i) The decay constant is the probability of decay of a nucleus per unit time (equivalently, the constant of proportionality between activity and the number of undecayed nuclei, \(A = \lambda N\)).

(a)(ii) \(t_{1/2} = 6.0 \times 3600 = 2.16\times10^{4}\) s. \(\lambda = \ln 2 / t_{1/2} = 0.693 / 2.16\times10^{4} = 3.21\times10^{-5}\ \text{s}^{-1}\), which is approximately \(3.2\times10^{-5}\ \text{s}^{-1}\).

(a)(iii) \(t = 20 \times 3600 = 7.2\times10^{4}\) s. \(A = A_0 e^{-\lambda t} = 4.8\times10^{5} \times e^{-(3.21\times10^{-5})(7.2\times10^{4})} = 4.8\times10^{5} \times e^{-2.31} = 4.8\times10^{5}\times 0.0992 = 4.8\times10^{4}\) Bq (4.76 x 10^4 Bq unrounded).

(b) \(I/I_0 = e^{-\mu x}\), so \(105/840 = e^{-15\mu}\). \(\ln(840/105) = \ln 8 = 2.079 = 15\mu\). \(\mu = 2.079/15 = 0.139\ \text{mm}^{-1}\) (0.14 mm^-1 to 2 s.f.).

(c) Indicative content: place the source and GM tube at a fixed separation with the lead sheet between them, source-detector distance and alignment unchanged throughout. Measure the thickness of each sheet with a micrometer (or vernier calipers), at several places, and take a mean. Measure the background count with the source removed or shielded, over a long time, and subtract it from every reading. For each thickness count for a long time (e.g. 60 s or more) so that the random uncertainty (about the square root of the count) is a small percentage, repeat and take a mean, and convert to a count rate. Use a range of thicknesses (including stacked sheets) so that the corrected rate stays clearly above the background. The half-life is much longer than the time of the experiment, so the source activity is effectively constant (or readings can be corrected for decay). Processing: \(C = C_0 e^{-\mu x}\) so \(\ln C = \ln C_0 - \mu x\). Plot \(\ln C\) against \(x\) with labelled axes and units; draw a best-fit straight line; gradient = \(-\mu\); the intercept gives \(\ln C_0\). Safety: handle the source with tongs, keep it as far from the body as practicable, point it away from people, keep exposure time short and return it to its lead-lined storage box immediately after use.

(a)(i) B1: probability (chance) of decay of a nucleus per unit time (AW; allow constant of proportionality in \(A = \lambda N\) with symbols defined). NOT 'rate of decay' alone. [1]

(a)(ii) C1: converts hours to seconds (\(6.0\times3600 = 2.16\times10^{4}\) s). A1: \(\lambda = \ln2/t_{1/2} = 3.2(1)\times10^{-5}\ \text{s}^{-1}\); the answer must be shown to at least 3 s.f. or the substitution clearly shown. Allow working in hours with a correct conversion at the end. [2]

(a)(iii) M1: use of \(A = A_0 e^{-\lambda t}\) with correct t (\(7.2\times10^{4}\) s), or \(A = A_0 (1/2)^{t/t_{1/2}}\) with \(t/t_{1/2} = 3.33\). A1: \(4.8\times10^{4}\) Bq. Allow \(4.7\times10^{4}\) to \(4.8\times10^{4}\) Bq. ECF from (a)(ii) for the value of \(\lambda\) only. [2]

(b) C1/M1: \(105/840 = e^{-15\mu}\) (or 840/105 = 8 used correctly). M1: takes natural logs correctly, \(\ln 8 = 2.08\), \(15\mu = 2.08\). A1: \(\mu = 0.139\ \text{mm}^{-1}\). Allow 0.14. Allow 139 m^-1 or 13.9 cm^-1 only if the unit is correct; unit given as required is mm^-1. Allow a half-value-thickness route: three half-values, so \(x_{1/2} = 5.0\) mm and \(\mu = \ln2/5.0 = 0.139\). Award full marks for a correct answer. [3]

(c)* Level 3 (5-6 marks): a clear, logical and complete description covering all three strands: measurements with at least two improvements to accuracy or precision (e.g. background subtraction, repeats or long counting times, fixed geometry, measuring thickness with a suitable instrument); correct linearisation and graph (ln C against x, gradient = -mu); and a valid safety precaution. Reasoning is well structured and the links between the method and the analysis are clear. Level 2 (3-4 marks): a description that covers two strands well or all three with some gaps; a method that would produce usable data but lacks some detail (e.g. the graph is mentioned without the log, or the background is mentioned without the corrected rate). Level 1 (1-2 marks): some relevant points with limited organisation, for example measuring count rate for different thicknesses, but with little about the improvement of accuracy or the analysis. 0 marks: no creditworthy response.
Indicative points (not a compulsory script): fixed source-detector distance and alignment; measure the background rate with the source removed and subtract it; count for a long time or repeat and average (random nature of decay, small percentage uncertainty); measure sheet thickness with a micrometer or calipers; use a range of thicknesses; the source activity is effectively constant over the time because the half-life is long (or correct for decay); plot ln(corrected count rate) against thickness; straight line with a negative gradient; \(\mu = -\) gradient; the intercept is \(\ln C_0\); safety: tongs, distance, short exposure, source not pointed at people, store in a lead-lined box. Allow plotting ln of the ratio \(C/C_0\). Ignore wearing gloves or goggles as the sole precaution. Allow a graph of C against x with an exponential fit, or a half-value-thickness method, only if \(\mu = \ln 2/x_{1/2}\) is used, but this may not score above Level 2 unless the averaging of several half-value thicknesses is described. [6]

Total: 14 marks.

(a)(i) \(\mu\) is the constant in \(I=I_0e^{-\mu x}\); it equals the fractional decrease in intensity per unit thickness of the material for a narrow beam.

(a)(ii) \(\ln(I_0/I)=\mu x\). \(\ln(640/160)=\ln 4=1.386\). \(\mu = 1.386/8.00 = 0.173\ \mathrm{mm^{-1}}\).

(a)(iii) For half intensity \(x_{1/2}=\ln 2/\mu = 0.693/0.1733 = 4.0\ \mathrm{mm}\).

(b) Percentage uncertainties: \(I_0\): 16/640 = 2.5%; \(I\): 8/160 = 5.0%. The ratio \(I_0/I\) has uncertainty 7.5%, so the absolute uncertainty in \(\ln 4\) is \(\approx 0.075\) (checking extremes: \(\ln(656/152)=1.462\), \(\ln(624/168)=1.312\)). This is \(0.075/1.386 = 5.4\%\). Uncertainty in \(x\): 0.02/8.00 = 0.25%. Total \(\approx 5.7\% \approx 6\%\). Absolute uncertainty \(= 0.057 \times 0.173 = 0.010\ \mathrm{mm^{-1}}\). So \(\mu = 0.173 \pm 0.010\ \mathrm{mm^{-1}}\).

(c) Indicative content: X-rays are produced by a tube and directed at the leg, with the detector (film or digital sensor) behind it. Absorption depends on \(\mu\), which is larger for bone (higher density and proton number, calcium) than for soft tissue, so bone transmits less and appears lighter/shows a fracture as a gap in the shadow. A narrow, collimated beam and a small source/focus give sharp images; an anti-scatter grid removes scattered X-rays that blur the image; placing the detector close to the leg reduces blurring. Contrast between tissues is improved by choosing a suitable X-ray photon energy; a contrast medium can be used for soft tissue. A metal (aluminium) filter removes low-energy X-rays, which would be absorbed in the skin without contributing to the image, reducing dose. A short exposure time, sensitive digital detector or intensifying screen lowers the dose. Shielding of other body areas and limiting the number of exposures also reduces risk.

(d) Intensity \(\propto 1/r^2\) (inverse square law): the same power is spread over a larger area as the beam diverges.

(a)(i) B1: fraction (proportion/percentage) of intensity absorbed/removed per unit thickness (AW); or constant in I = I0 e^(-mu x) with the symbols defined. Allow 'depends on material and photon energy'. NOT just 'how much is absorbed'. [1]

(a)(ii) M1: ln(I0/I) = mu x or ln(160/640) = -mu x (ln 4 = 1.386 seen). A1: mu = 0.173 mm^-1 (allow 0.17; allow 173 m^-1 with unit). Allow a 1-mark answer if the unit is missing or wrong with correct value. [2]

(a)(iii) C1: x = ln 2 / mu (allow ln 2 = mu x, or I/I0 = 0.5 used). A1: 4.0 mm (ECF from (a)(ii); allow 4 mm; also allow answer of exactly 4.0 mm from 2 x 8/4 reasoning, i.e. 640 -> 320 -> 160 two half-thicknesses). [2]

(b) M1: percentage uncertainties of I0 (2.5%) and I (5.0%) added (7.5%) or extreme values of ln (1.312 and 1.462) used. A1: percentage uncertainty in ln(I0/I) = 5.4% (allow 5% to 6%). A1: total percentage uncertainty in mu approx 5.7% (allow 6%, allow 5.5% to 6%) and mu = 0.173 +/- 0.010 mm^-1 (allow +/-0.01). Do not penalise 0.17 +/- 0.01. Allow ECF from (a)(ii) for the value of mu only. Do NOT credit adding the percentage uncertainty of ln 4 as 7.5% directly (max 1 mark). [3]

(c)* Level of response, 6 marks. Indicative scientific points: (1) X-ray photons produced by tube, passed through leg to detector; (2) bone has greater mu / absorbs more than soft tissue (higher density / proton number); (3) differential absorption produces contrast / shadow image; (4) collimated narrow beam or small source gives sharper image; (5) anti-scatter grid or detector close to leg to reduce blurring; (6) suitable photon energy or contrast medium to improve contrast; (7) aluminium filter removes low-energy X-rays so reducing patient dose; (8) short exposure, sensitive detector or intensifying screen reduces dose; (9) shielding or limiting number of exposures.
5-6 marks: Describes how the image is formed AND gives explained methods for improving both image quality and patient safety, linking each to the physics (e.g. mu, absorption, scattering, filtering). Reasoning is logical, complete and well structured, with at least two points supported for each of quality and safety.
3-4 marks: Describes image formation with some explanation, and gives measures for improving quality and/or safety. Some links are made to the physics but the explanation is incomplete, or one strand (quality or safety) is missing or only stated.
1-2 marks: Makes isolated relevant statements (e.g. bone absorbs more X-rays; a filter is used) with little or no explanation or logical structure.
0 marks: No response or no creditworthy response. [6]

(d) B1: inverse square law / intensity proportional to 1/r^2 / energy (power) spread over increasing area as the beam spreads out (AW). Ignore 'attenuation'. [1]

Total: 15 marks.