Pearson Edexcel International A Level Physics: Unit 6: analysing data, log graphs and uncertainties (YPH11) – Practice Questions

These are original questions written for Marlbridge, for revision and practice on this content. They are not reproduced past-paper questions, and they do not replicate the exam’s exact structure, question count or mark tariffs – examination boards hold copyright in their own papers. Use these alongside the official past papers from your board or school.

These questions cover section 6.5 Analysis of Unit 6: Practical Skills in Physics II in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (YPH11), Issue 3. Unit 6 is IA2 and has no tiers. Learn the ideas from the study guide or the revision notes first. More links are at the end.

Questions

1. A student measures a potential difference of 4.8 V and a current of 0.36 A. Calculate the resistance, giving your answer to a suitable number of significant figures with its unit. [2]

2. (a) A vernier caliper has a resolution of 0.02 mm and gives a single reading of 12.34 mm. Find the percentage uncertainty. [2] (b) Five timings of a swing are 1.52 s, 1.60 s, 1.47 s, 1.55 s and 1.56 s. Find the mean and the percentage uncertainty using the half range. [2]

3. A spring has length 14.2 cm ± 0.1 cm unloaded and 21.7 cm ± 0.1 cm when loaded. Find the extension, its absolute uncertainty and its percentage uncertainty. [3]

4. A solid cylinder has mass 62.4 g ± 0.1 g, diameter 2.15 cm ± 0.01 cm and height 6.80 cm ± 0.05 cm. Calculate its density in kg m⁻³ and the uncertainty in the density. [5]

5. A student measures a standard 100.0 Ω resistor five times in two ways. Set A gives 98.1, 98.3, 98.2, 98.2 and 98.1 Ω. Set B gives 99.0, 101.5, 97.8, 102.3 and 100.4 Ω. (a) Compare the sets using the words precise and accurate. [2] (b) A voltmeter can be used on a 0 to 2 V range or a 0 to 200 V range. Explain which is more sensitive for measuring a 0.35 V signal. [2]

6. A small wind turbine gives power P (watts) at wind speed v (m s⁻¹): v = 4.0, 6.0, 8.0, 10.0, 12.0 and P = 35, 118, 280, 545, 940. It is suggested that P = kvⁿ. (a) State what to plot to test this. [1] (b) Using the points (6.0, 118) and (12.0, 940), find n. [2] (c) Find k, with its unit. [2] (d) State the relationship. [1]

7. The corrected count rate A of a radioactive sample is 950, 672, 475, 336, 238 and 168 counts per minute at 0, 2, 4, 6, 8 and 10 h. A graph of ln A against t has a best-fit line through (0, 6.86) and (10, 5.12). (a) Explain why ln A is plotted against t. [1] (b) Find the gradient and the decay constant. [2] (c) Find the half-life. [2]

8. A graph of T² against length l for a pendulum should pass through the origin. The best-fit line crosses the vertical axis at +0.05 s². (a) State what this suggests and give a possible cause. [2] (b) One point lies well off the line. State what you would do. [1] (c) Why is a large triangle used to find the gradient? [1]

9. A ball falls from rest through h = 1.150 m ± 0.002 m. Five timings are 0.489 s, 0.468 s, 0.497 s, 0.481 s and 0.485 s. Use g = 2h/t². (a) Find the mean time and its percentage uncertainty. [2] (b) Calculate g. [1] (c) Find the percentage and absolute uncertainty in g. [3] (d) Compare your result with 9.81 m s⁻² and comment on its precision. [2]

10. A student times a ball bearing falling through oil in a tall cylinder with a hand stopwatch over 20 cm marked from the top, with one reading per ball and the oil warming in the student’s hands. Suggest three realistic modifications to improve the experiment, each with its effect. [6]

11. The terminal speed v (cm s⁻¹) of spheres of radius r (mm) in a liquid is 0.85, 1.88, 3.40, 5.18 and 7.62 for r = 1.0, 1.5, 2.0, 2.5 and 3.0. Test v = krⁿ using the points (r = 1.0, v = 0.85) and (r = 3.0, v = 7.62). (a) Calculate log₁₀ r and log₁₀ v for these two points. [2] (b) Find n. [2] (c) Find k. [1] (d) Predict v at r = 4.0 mm. [2]

12. A table gives current I/A as 0.25, 0.3, 0.350 and 0.40 for p.d. V/V of 1.2, 1.5, 1.8 and 2.0, with resistance R/Ω as 4.8, 5.000, 5.142857 and 5. (a) The ammeter has a resolution of 0.01 A. Identify and correct the faulty current entries. [2] (b) Rewrite the resistance column to a suitable number of significant figures. [2]

Answers

1. R = V/I = 4.8/0.36 = 13.33 [1]. Both data have 2 significant figures, so R = 13 Ω [1]. [2] Examiner insight: Show the unrounded value, then round; an answer with no unit or too many significant figures can lose credit.

2. (a) Uncertainty = half the resolution = 0.01 mm [1]; percentage = 0.01/12.34 × 100 = 0.081% [1]. (b) Mean = 1.54 s; half range = (1.60 - 1.47)/2 = 0.065 s [1]; percentage = 0.065/1.54 × 100 = 4.2% [1]. [4] Examiner insight: Single reading means half the resolution; repeats mean half the range. Using 0.02 mm in part (a) is the usual slip.

3. Extension = 21.7 - 14.2 = 7.5 cm [1]. Absolute uncertainties add: 0.1 + 0.1 = ±0.2 cm [1]. Percentage = 0.2/7.5 × 100 = 2.7% [1]. [3] Examiner insight: A difference adds absolute uncertainties; adding the two percentage uncertainties instead gives a wrong figure and is a common error.

4. ρ = 4m/(πd²h) = 4 × 62.4/(π × 2.15² × 6.80) = 2.53 g cm⁻³ [1] = 2530 kg m⁻³ [1]. Percentages: mass 0.16%, diameter 0.47%, height 0.74% [1]. Total = 0.16 + 2 × 0.47 + 0.74 = 1.8% [1]. Absolute = 2530 × 1.8% = ±46 kg m⁻³ (about ±50) [1]. [5] Examiner insight: The factor of 2 on the diameter is the step most often missed; set out each percentage on its own line so that credit for method is visible.

5. (a) Set A has a small spread so is precise but its mean of 98.2 Ω is not close to 100.0 Ω so it is not accurate, which points to a systematic error [1]. Set B has a large spread so is less precise, but its mean of 100.2 Ω is close to the true value, so it is more accurate [1]. (b) The 0 to 2 V range is more sensitive [1] because a given 0.35 V gives a larger deflection or finer reading, with a smaller smallest division, than on the 200 V range [1]. [4] Examiner insight: Link each term to the evidence: spread for precision, closeness to 100.0 Ω for accuracy. A bare label scores less than a label plus the reason.

6. (a) log P against log v [1]. (b) log 6.0 = 0.778 and log 118 = 2.072; log 12.0 = 1.079 and log 940 = 2.973 [1]. Gradient = (2.973 - 2.072)/(1.079 - 0.778) = 2.99, so n = 3.0 [1]. (c) log k = 2.072 - 3.0 × 0.778 = -0.262 [1], so k = 10^-0.262 = 0.55 W s³ m⁻³ [1]. (d) P = 0.55 v³ [1]. [6] Examiner insight: Take logs of the numbers only. The unit of k comes from P = kv³.

7. (a) A = A₀e^(-λt), so ln A = ln A₀ - λt, a straight line with gradient -λ [1]. (b) Gradient = (5.12 - 6.86)/10 = -0.174 h⁻¹ [1]; so λ = 0.174 h⁻¹ [1]. (c) T½ = ln 2/λ = 0.693/0.174 [1] = 4.0 h [1]. [5] Examiner insight: Choose points on the line that are far apart. State the sign of the gradient, then say that λ is its magnitude.

8. (a) A positive intercept suggests a systematic error [1]; for example, the length measured to the top of the bob rather than to its centre [1]. (b) Check and repeat that reading [1]. (c) A large triangle reduces the percentage error in reading the points from the graph [1]. [4] Examiner insight: Name the cause in terms of the quantity measured. “Human error” is too vague to be credited.

9. (a) Mean = 0.484 s [1]; half range = (0.497 - 0.468)/2 = 0.0145 s, which is 0.0145/0.484 = 3.0% [1]. (b) g = 2 × 1.150/0.484² = 9.82 m s⁻² [1]. (c) Percentage in h = 0.002/1.150 × 100 = 0.17% [1]. Total = 0.17 + 2 × 3.0 = 6.2% [1]. Absolute = 9.82 × 6.2% = ±0.61 m s⁻² [1]. (d) Range = 9.2 to 10.4 m s⁻², which includes 9.81, so the result is accurate within its uncertainty [1]. The percentage uncertainty is above 5%, so the timing is not very precise [1]. [8] Examiner insight: Compare the accepted value with the range from your own uncertainty. A conclusion without a number from the working is not complete.

10. Any three changes, each with its effect. Use light gates or video with a scale [1] to remove reaction-time error [1]. Start the timed section lower down [1] so that the ball has reached terminal speed [1]. Keep the oil in a water bath [1] so that the viscosity stays constant [1]. Other valid changes: repeat and average each radius; use a micrometer for the radius. [6] Examiner insight: Pair each change with its benefit. A list of apparatus names with no effect gains less.

11. (a) For r = 1.0: log r = 0.000 and log v = -0.071; for r = 3.0: log r = 0.477 and log v = 0.882 [2]. (b) Gradient = (0.882 + 0.071)/0.477 [1] = n = 2.0 [1]. (c) The intercept at log r = 0 is -0.071, so k = 10^-0.071 = 0.85 [1]. (d) v = 0.85 × 4.0² [1] = 13.6, so 14 cm s⁻¹ [1]. [7] Examiner insight: A prediction should give your law and the substituted value; round to the precision of the data.

12. (a) Entries 0.3 and 0.350 [1] should be 0.30 and 0.35, since readings are recorded to the resolution of the instrument [1]. (b) Two significant figures, to match the data [1]: 4.8, 5.0, 5.1, 5.0 Ω [1]. [4] Examiner insight: Raw readings share decimal places set by the instrument; a calculated result follows the least precise input.

Where marks are usually lost

• Reporting a long calculator display instead of rounding.
• Using the whole resolution, or the range, as the uncertainty when half is required.
• Adding absolute uncertainties for a product, or adding percentages for a difference.
• Not multiplying a percentage uncertainty by the power for a squared or cubed quantity.
• Including the unit inside the logarithm, or labelling a log axis with no unit information.
• Taking a gradient over a short interval or from data points rather than the line.
• Using precise and accurate interchangeably.
• Suggesting improvements that are vague or unrealistic.

Next steps

• Revision notes for this topic
• Study guide for this topic
• Edexcel International A Level Physics hub
• Printable checklist
• All free 10-minute diagnostics
• Book a free trial class

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics specification (YPH11), Issue 3, Pearson Education Limited, section 6.5 Analysis.