Pearson Edexcel International A Level Physics: Unit 3: processing results, 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 3.5 Processing Results of Unit 3: Practical Skills in Physics I, an International AS (IAS) unit, in the Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics (YPH11) specification, Issue 3. As the specification does not expect compounding of percentage uncertainties, none is asked for. Revise with the study guide or the revision notes. See also the Edexcel A Level Physics hub, the printable checklist and the diagnostics.

Questions

1. A voltmeter reads 4.8 V and an ammeter reads 0.250 A for the same component. Calculate its resistance, giving your answer to a suitable number of significant figures. [2]

2. Calculate the percentage uncertainty in each reading, using half the resolution of the instrument.

(a) A rule with 1 mm divisions is used to measure a length of 18.5 cm. [2] (b) A micrometer with resolution 0.01 mm is used to measure a wire of diameter 0.82 mm. [2]

3. A student times 10 oscillations of a mass on a string. Five repeats, in seconds, are 14.8, 15.1, 14.9, 15.3 and 14.7.

(a) Calculate the mean time for 10 oscillations and the period. [2] (b) Calculate the percentage uncertainty in the mean time, using the half range. [2]

4. A student plots extension against force on a grid 16 cm wide and 12 cm tall. The force runs from 0 to 9.0 N and the extension from 0 to 126 mm.

(a) State a suitable scale for each axis, with extension on the horizontal axis. [2] (b) The student suggests 7 mm per centimetre for the extension axis. Give two reasons why this is a poor choice. [2]

5. A spring is loaded and the extension x is measured. The best-fit line of force F (in N) against x (in mm) passes through the points (10, 2.0) and (60, 12.0).

(a) Calculate the gradient of the line, with its unit. [2] (b) Convert the gradient to N m⁻¹. [1] (c) Explain why points this far apart were chosen. [1] (d) The line also passes through (0, 0). State what this shows about F and x. [1]

6. A ball is released from rest and the distance fallen h is measured for different fall times t.

t / s	0.20	0.30	0.40	0.50	0.60
h / m	0.22	0.47	0.81	1.25	1.79

(a) Calculate t² for each time, to a suitable number of significant figures. [1] (b) Explain why a graph of h against t² is expected to be a straight line. [1] (c) The best-fit line of h against t² passes through (0.05, 0.27) and (0.35, 1.74). Determine the gradient and use it to find the acceleration of free fall. [3] (d) The line meets the h-axis at +0.025 m. Suggest one cause. [1]

7. A student measures the terminal potential difference V of a cell for several currents I. The best-fit line of V against I passes through (0, 1.52) and (0.50, 0.82), with V in volts and I in amperes.

(a) State what the intercept on the V-axis represents and give its value. [2] (b) Calculate the gradient, with its unit. [2] (c) Use V = ε − Ir to find the internal resistance of the cell. [1] (d) A classmate says the line should be forced through the origin. Explain why this is wrong. [1]

8. A micrometer is closed with nothing between the jaws and reads +0.02 mm. A wire is then measured and the reading is 0.48 mm.

(a) Name the type of error and give the correct diameter. [2] (b) Explain why repeating the reading and taking a mean does not remove this error. [1] (c) Give one source of random error when timing by hand, and one way to reduce its effect. [1]

9. A student investigates how the period of a pendulum depends on its length. Each period is found by timing a single oscillation with a hand stopwatch. The student starts and stops the watch as the bob passes any convenient point. Only three lengths, between 0.30 m and 0.40 m, are used. The length is measured with a rule held by hand. Suggest five realistic modifications, each with the reason it helps. [5]

10. A student measures the Young modulus of a wire of length 2.00 m. Five micrometer readings of its diameter, in mm, are 0.46, 0.47, 0.45, 0.46 and 0.46. The best-fit line of extension e against load F passes through (0, 0) and (12.0 N, 0.75 mm).

(a) Calculate the mean diameter, the half range and the percentage uncertainty in the diameter. [3] (b) Calculate the cross-sectional area of the wire. [2] (c) Calculate the gradient of the line in m N⁻¹. [1] (d) Use E = FL / (Ae) to find the Young modulus. [2] (e) The metre rule used for the length has 1 mm divisions. Compare its percentage uncertainty with that of the diameter and suggest one improvement. [2]

Answers

1. R = V / I = 4.8 / 0.250 = 19.2 Ω [1]. The voltage has 2 s.f., so the answer is 19 Ω [1]. [2] Examiner insight: Give the final answer to the s.f. of the least precise value, and keep the extra digit only in the working.

2. (a) Uncertainty = 0.5 mm [1]; 0.5 / 185 × 100 = 0.27% [1]. (b) Uncertainty = 0.005 mm [1]; 0.005 / 0.82 × 100 = 0.61% [1]. Examiner insight: Write the uncertainty first and then the percentage. Using the whole resolution, or forgetting to convert 18.5 cm to 185 mm, is a slip that changes the answer.

3. (a) Mean = 74.8 / 5 = 14.96 s [1]. Period = 14.96 / 10 = 1.50 s [1]. (b) Half range = (15.3 − 14.7) / 2 = 0.3 s [1]. 0.3 / 14.96 × 100 = 2.0% [1]. Examiner insight: Show the largest and smallest values you used, then divide by 2.

4. (a) Extension: 10 mm per 1 cm, so 126 mm takes 12.6 cm of the 16 cm [1]. Force: 1 N per 1 cm, so 9.0 N takes 9.0 cm of the 12 cm [1]. (b) 126 / 7 = 18 cm, which is longer than the 16 cm grid, so the points do not fit [1]. Steps of 7 are awkward to subdivide and read, so plotting is slow and error-prone [1]. Examiner insight: A scale answer is stronger when you state how much of the grid the data fill.

5. (a) Gradient = (12.0 − 2.0) / (60 − 10) [1] = 0.200 N mm⁻¹ [1]. (b) 0.200 N per 10⁻³ m = 200 N m⁻¹ [1]. (c) A long baseline means a reading error of a fraction of a small square is a small fraction of the change, so the gradient is more reliable [1]. (d) The line through the origin shows F is directly proportional to x [1]. Examiner insight: Include the unit in the gradient before converting. Converting mm to m in the wrong direction, giving 0.0002, is an easy error to make.

6. (a) 0.040, 0.090, 0.16, 0.25, 0.36 s² (2 s.f.) [1]. (b) From rest, h = ½gt², which has the form y = mx with y = h and x = t² [1]. (c) Gradient = (1.74 − 0.27) / (0.35 − 0.05) [1] = 4.90 m s⁻² [1]. The gradient is g/2, so g = 2 × 4.90 = 9.80 m s⁻² [1]. (d) A systematic error, for example the timer started a little after release, or h was measured from a point 2.5 cm above the true release point [1]. Examiner insight: After finding the gradient, say what it equals. Stopping at 4.90 without doubling it leaves the answer incomplete.

7. (a) The intercept is the emf of the cell [1], 1.52 V [1]. (b) Gradient = (0.82 − 1.52) / (0.50 − 0) [1] = −1.40 V A⁻¹ (that is, −1.40 Ω) [1]. (c) V = ε − Ir, so the gradient is −r and r = 1.40 Ω [1]. (d) When I = 0 the terminal pd equals the emf, which is not zero [1]. Examiner insight: Keep the minus sign in the gradient, then say that r is positive.

8. (a) A systematic (zero) error [1]; correct diameter = 0.48 − 0.02 = 0.46 mm [1]. (b) Each reading is out by the same offset, so the mean carries that offset as well [1]. (c) Reaction time varies from one timing to the next [1]; repeating and taking a mean, or using a light gate, reduces its effect. Credit one source with a matching remedy. Examiner insight: Part (c) needs a source and a remedy that fit each other.

9. Any five of these, each with its reason, one mark each:

• Time 20 oscillations and divide by 20, so the reaction-time uncertainty is a smaller percentage of the total [1].
• Use a fiducial marker at the centre of the swing as the start and stop point, so the same point is used each time [1].
• Repeat each timing and take a mean, to reduce random error [1].
• Use many more lengths over a wider range, such as 0.20 m to 1.00 m, so a trend is clear on the graph [1].
• Clamp the rule vertically, check with a set square, and read at eye level to reduce parallax [1]. Examiner insight: “Use more accurate equipment” gains nothing. Name the change, and say which error it reduces.

10. (a) Mean = 2.30 / 5 = 0.460 mm [1]; half range = (0.47 − 0.45) / 2 = 0.010 mm [1]; 0.010 / 0.460 × 100 = 2.2% [1]. (b) A = π × (0.230 × 10⁻³)² [1] = 1.66 × 10⁻⁷ m² [1]. (c) Gradient = 0.75 × 10⁻³ / 12.0 = 6.25 × 10⁻⁵ m N⁻¹ [1]. (d) E = FL / (Ae), so e / F = L / (AE) and E = L / (A × gradient) [1] = 2.00 / (1.66 × 10⁻⁷ × 6.25 × 10⁻⁵) = 1.93 × 10¹¹ Pa [1]. (e) Rule: 0.5 / 2000 × 100 = 0.025%, far smaller than the 2.2% in the diameter [1]. Improve the diameter by measuring at several points along the wire and at right angles to each other, and averaging [1]. Examiner insight: Convert mm to m before using the formula. In part (e) a comparison of two figures is more convincing than a statement that one is larger.

Where marks are usually lost

• Giving a final answer with more significant figures than the data support.
• Dividing the range by 2 in the wrong place, or forgetting to halve it at all.
• Using the resolution, not half the resolution, as the uncertainty of one reading.
• Mixing millimetres and metres in gradients and areas.
• Gradient triangles that are small, or that use plotted points not on the line.
• Stopping at the gradient and not equating it to the physical quantity.
• Calling an offset or intercept random error when it is systematic.
• Vague improvements that name no instrument and no error. Missing units on axes and gradients.

Next steps

• Re-read the revision notes and the study guide.
• Go to the Edexcel A Level Physics hub and tick off the printable checklist.
• Take one of the free 10-minute diagnostics.
• Book a free trial class on the trial page.

Official syllabus

Pearson Edexcel International Advanced Subsidiary/Advanced Level in Physics (YPH11), Specification, Issue 3, published by Pearson Education Limited. Section 3.5 Processing Results, within Unit 3: Practical Skills in Physics I.