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Cambridge practical and extended reasoning

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AI-assisted English content review; independent teacher approval pending.

Practice scope and review

Original, AI-authored English practice with worked guidance. Selected topics only; independent teacher review is pending. This is not a complete syllabus, official past-paper collection or predicted exam score. Check your own exam-board specification: these concepts are not certified against a particular board or exam year.

91 original questions across 15 selected topic areas. No account or payment required to practise.

Cambridge International AS & A Level Physics (9702), examination year 2027: AI-assisted conceptual alignment to official sources, not teacher or exam-board certification. Syllabus mapping and remaining coverage.

4 questions in this topic

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Original multi-stage supplied-data practice. Partial written support only, not complete syllabus coverage, official results or practical certification.

Subtopics in this topic (4)
  • Resistivity, intercept and correlated errors
  • Log-linear discharge and model failure
  • Paired heating and energy-loss models
  • Light-gate acceleration and uncertainty conventions

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Question 1Written analysis of fictional wire measurements: length L / m = 0.20, 0.40, 0.60, 0.80 and resistance R / ohm = 1.10, 1.90, 2.70, 3.50. Diameter is 0.400 +/- 0.010 mm. Use R = rho L/A + Rcontact, A = pi d^2/4. Treat lengths and gradient as exact for the diameter-only bound. Find gradient, intercept and resistivity; estimate the first-order worst-case percentage contribution from diameter. Explain why forcing the graph through zero is unjustified, propose controls and distinguish a common calibration error from random scatter. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf5 indicative marks
Calculate, analyse and evaluate5 indicative marksCalculator allowed

Written analysis of fictional wire measurements: length L / m = 0.20, 0.40, 0.60, 0.80 and resistance R / ohm = 1.10, 1.90, 2.70, 3.50. Diameter is 0.400 +/- 0.010 mm. Use R = rho L/A + Rcontact, A = pi d^2/4. Treat lengths and gradient as exact for the diameter-only bound. Find gradient, intercept and resistivity; estimate the first-order worst-case percentage contribution from diameter. Explain why forcing the graph through zero is unjustified, propose controls and distinguish a common calibration error from random scatter. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf

Cambridge 2027: Partial conceptual match

Resistivity, intercept and correlated errors: selected written evidence only. Gradient/intercept, SI area and rho recomputed; diameter bound distinguished from NIST standard uncertainty; offset inference conditional. Science practical-N and computing rN rows are internal identifiers, not official numbered leaf syllabus codes. Aggregate rows are only partially supported. Actual apparatus use, candidate coding execution and practical attainment are not demonstrated here. Independent teacher review remains required.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Use a long gradient triangle: (3.50 - 1.10)/(0.80 - 0.20) = 4.00 ohm/m. Substitution at 0.20 m gives 1.10 - 4.00(0.20) = 0.30 ohm. The middle rows lie on that line, but ideal fictional alignment does not show real apparatus is exact.
  2. Convert diameter before squaring: d = 4.00e-4 m. Multiplying the gradient by pi d2d^{2}/4 gives 5.0265e-7 ohm m, rounded to 5.03e-7. Since rho is proportional to d2d^{2}, the first-order bound is 2(0.010/0.400) x 100 = 5.0%. This is a stipulated bound contribution, not a standard uncertainty or a complete budget.
  3. A zero-constrained fit absorbs the offset into its slope and biases rho. A separate lead-resistance check is preferable to declaring the intercept certainly a contact effect. Keep temperature stable to avoid resistance drift and sample diameters in several orientations. Repeats estimate scatter; a shared micrometer scale bias survives averaging and is correlated across all area estimates. Actual instrument handling remains unobserved.

Self-marking points

View marking points
  1. Gradient is 4.00 ohm/m and intercept 0.30 ohm.
  2. Area is 1.2566e-7 m^2; rho = 5.03e-7 ohm m (3 significant figures).
  3. Diameter-only first-order worst-case contribution is 5.0%, not 2.5%.
  4. The intercept is consistent with lead/contact resistance under this model, not proof of its cause.
  5. Control wire temperature using teacher-approved low-energy equipment; inspect diameter along the wire and calibration separately.

Examiner tip

Indicative original worked practice, independent teacher approval pending. Explain assumptions and alternatives; do not equate a correct written answer with observed practical performance.

Question 2Fictional capacitor records give t / s = 0, 20, 40 and V / V = 6.00, 3.64, 2.21. The model is V = V0 exp(-t/RC); R = 100 +/- 2 kilohm. Find the slope of ln(V/1 V) against time using endpoints, RC and C. Treat slope uncertainty as negligible only for the resistance contribution. A later 80 s reading is 1.20 V. Predict the model reading, evaluate the discrepancy without inventing significance, and plan controls/checks for an approved supervised measurement. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf5 indicative marks
Calculate, analyse and evaluate5 indicative marksCalculator allowed

Fictional capacitor records give t / s = 0, 20, 40 and V / V = 6.00, 3.64, 2.21. The model is V = V0 exp(-t/RC); R = 100 +/- 2 kilohm. Find the slope of ln(V/1 V) against time using endpoints, RC and C. Treat slope uncertainty as negligible only for the resistance contribution. A later 80 s reading is 1.20 V. Predict the model reading, evaluate the discrepancy without inventing significance, and plan controls/checks for an approved supervised measurement. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf

Cambridge 2027: Partial conceptual match

Log-linear discharge and model failure: selected written evidence only. Natural-log slope, inverse slope, SI capacitance and extrapolated residual independently checked; no uncertainty invented. Science practical-N and computing rN rows are internal identifiers, not official numbered leaf syllabus codes. Aggregate rows are only partially supported. Actual apparatus use, candidate coding execution and practical attainment are not demonstrated here. Independent teacher review remains required.

AI-assisted placement. Independent teacher approval pending.

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Worked guidance

  1. Linearise with a dimensionless voltage ratio: m = ln(2.21/6.00)/40 = -0.0249692 s−1s^{-1}. Thus tau = -1/m = 40.049 s. The 20 s point is approximately consistent, but three records cannot establish a universal exponential law.
  2. Use R = 100000 ohm: C = tau/R = 0.00040049 F, or about 400 microfarad to three significant figures. Holding tau fixed gives |delta C/C| approximately |delta R/R| = 2%; the full uncertainty would also include slope and model errors.
  3. V(80) = 6.00 exp(2 ln(2.21/6.00)) = 0.8140 V. A positive zero offset could flatten late readings; leakage or loading could instead alter tau, so inspect residuals over a broader range and check a zero-input record. Repeat from the same initial voltage and record meter input resistance. Do not infer a unique fault or direct anyone to handle a charged capacitor from this written exercise.

Self-marking points

View marking points
  1. Endpoint log slope is approximately -0.024969 s^-1; RC = 40.05 s.
  2. Capacitance is 4.00e-4 F, about 400 microfarad, with 2% resistance-only first-order uncertainty contribution.
  3. The predicted 80 s voltage is 0.814 V; the observed 1.20 V is about 0.386 V higher.
  4. No voltage uncertainty is supplied, so formal rejection or statistical significance cannot be asserted.
  5. Check meter loading, zero offset and timing; use supervised approved low-voltage apparatus and safe discharge procedures.

Examiner tip

Indicative original worked practice, independent teacher approval pending. Explain assumptions and alternatives; do not equate a correct written answer with observed practical performance.

Question 3Supplied fictional heating records for the same 0.200 kg sample give temperature-rise rates 0.030 K/s at electrical power 30.0 W and 0.050 K/s at 50.0 W. Use P = (mc + Ccontainer)r + H, with container heat capacity 100 J/K and constant heat-loss power H at the matched temperature. Determine c and H. Evaluate the claim that H = 0 proves perfect insulation. Predict the rate at 40.0 W and propose a control that tests the constant-loss assumption without performing an experiment. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf5 indicative marks
Calculate, analyse and evaluate5 indicative marksCalculator allowed

Supplied fictional heating records for the same 0.200 kg sample give temperature-rise rates 0.030 K/s at electrical power 30.0 W and 0.050 K/s at 50.0 W. Use P = (mc + Ccontainer)r + H, with container heat capacity 100 J/K and constant heat-loss power H at the matched temperature. Determine c and H. Evaluate the claim that H = 0 proves perfect insulation. Predict the rate at 40.0 W and propose a control that tests the constant-loss assumption without performing an experiment. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf

Cambridge 2027: Partial conceptual match

Paired heating and energy-loss models: selected written evidence only. Two-equation elimination, container correction, intercept and interpolation checked; two-point identifiability separated from validation. Science practical-N and computing rN rows are internal identifiers, not official numbered leaf syllabus codes. Aggregate rows are only partially supported. Actual apparatus use, candidate coding execution and practical attainment are not demonstrated here. Independent teacher review remains required.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Subtract the two equations to eliminate the assumed common H: (50.0 - 30.0)/(0.050 - 0.030) = 1000 J/K. Remove the container contribution before dividing by sample mass: c = (1000 - 100)/0.200 = 4500 J/(kg K). Ignoring the container would give 5000 and overestimate c.
  2. Substitute back: H = 30.0 - 1000(0.030) = 0 W, then r = (40.0 - 0)/1000 = 0.040 K/s. Keep working precision but report c about 4.5e3 because the supplied rise rates have two significant figures.
  3. The constant-loss model is an assumption at matched temperatures, not a fact proved by its own fit. Loss commonly changes as sample temperature rises. Compare initial slopes at the same sample/environment temperatures with the same insulation and mixing; extra power settings and cooling records could expose curvature or an offset. Temperature lag and uncertain electrical power can mimic a changed heat capacity. A teacher must approve thermal and electrical risks; no heating is requested here.

Self-marking points

View marking points
  1. Total effective heat capacity is 1000 J/K.
  2. Specific heat capacity is 4500 J/(kg K), or 4.5e3 at two significant figures.
  3. Fitted H is 0 W and predicted rate at 40.0 W is 0.040 K/s.
  4. Two points determine two model parameters exactly; a zero residual or intercept is not evidence of perfect insulation.
  5. Compare matched temperatures and stirring conditions and collect additional independent power settings; supervise any approved heated apparatus.

Examiner tip

Indicative original worked practice, independent teacher approval pending. Explain assumptions and alternatives; do not equate a correct written answer with observed practical performance.

Question 4A fictional trolley carries a 0.0500 m flag. Light-gate interruption times are 0.100 s and 0.0500 s; the midpoints of those intervals are 0.400 s apart. Approximate each instantaneous speed by flag length/interruption time and calculate acceleration. For a separate first-gate speed estimate, standard uncertainties are u(L)=0.0005 m and u(t)=0.002 s, independent; use u(v)/v=sqrt((u(L)/L)^2+(u(t)/t)^2). Compare with a first-order bound if those same numbers were maximum errors instead. Explain finite-flag bias and why two speeds share a length error. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf5 indicative marks
Calculate, analyse and evaluate5 indicative marksCalculator allowed

A fictional trolley carries a 0.0500 m flag. Light-gate interruption times are 0.100 s and 0.0500 s; the midpoints of those intervals are 0.400 s apart. Approximate each instantaneous speed by flag length/interruption time and calculate acceleration. For a separate first-gate speed estimate, standard uncertainties are u(L)=0.0005 m and u(t)=0.002 s, independent; use u(v)/v=sqrt((u(L)/L)^2+(u(t)/t)^2). Compare with a first-order bound if those same numbers were maximum errors instead. Explain finite-flag bias and why two speeds share a length error. Give a connected worked response with calculations, justified conclusions and limitations. All datasets are original fictional practice, not official experiment results. Written/supplied-data work only; any actual laboratory or field activity requires teacher approval, risk assessment and supervision. No human trials. Primary reading: https://www.cambridgeinternational.org/Images/664565-2025-2027-syllabus.pdf

Cambridge 2027: Partial conceptual match

Light-gate acceleration and uncertainty conventions: selected written evidence only. Speeds, acceleration, RSS standard uncertainty and alternative linear bound checked separately, retaining shared-length correlation. Science practical-N and computing rN rows are internal identifiers, not official numbered leaf syllabus codes. Aggregate rows are only partially supported. Actual apparatus use, candidate coding execution and practical attainment are not demonstrated here. Independent teacher review remains required.

AI-assisted placement. Independent teacher approval pending.

Show indicative guidance

Worked guidance

  1. Compute speeds with the same SI flag length: 0.0500/0.100 = 0.500 and 0.0500/0.0500 = 1.00 m/s. Using the supplied midpoint separation gives a = (1.00 - 0.500)/0.400 = 1.25 m/s^2. A constant-acceleration assumption makes midpoint speeds appropriate; without it this is an interval estimate.
  2. For the stipulated independent standard uncertainties, the relative value is sqrt(0.01^2 + 0.02^2) = 0.0223607, giving u(v) = 0.01118 m/s. Report roughly 0.500 +/- 0.011 m/s as a standard uncertainty. Adding instead gives 3.0% and 0.015 m/s only for the alternative first-order worst-case interpretation; these are not interchangeable confidence statements.
  3. Reduce finite-interval bias by suitable shorter flags if timing resolution permits; compare estimates at different flag lengths. Check alignment, repeat passes and keep the track configuration fixed. A single scale error in L affects both speeds in the same direction; a full acceleration budget must preserve that shared dependence. Any actual trolley setup needs teacher approval, secure equipment and a stop zone; the records demonstrate no handling.

Self-marking points

View marking points
  1. Gate speeds are 0.500 and 1.00 m/s; acceleration is 1.25 m/s^2.
  2. First-gate combined standard uncertainty is 0.0112 m/s, or 2.24%.
  3. The alternative first-order maximum-error bound is 0.015 m/s, or 3.0%, under the stated different interpretation.
  4. A flag time measures average speed over a finite interval, not an exactly instantaneous reading.
  5. The common flag length correlates the two inferred speeds; it must not be treated as two independent length errors.

Examiner tip

Indicative original worked practice, independent teacher approval pending. Explain assumptions and alternatives; do not equate a correct written answer with observed practical performance.

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