Measuring well: SI units, uncertainty, significant figures and the two kinds of error
By the end of this lesson you will be able to
- State the seven SI base units and use prefixes from nano to giga, converting between units with powers of ten.
- Select the instrument with the right resolution for a measurement and read it correctly — at eye level, to the bottom of the meniscus, allowing for a zero error.
- Distinguish precision, accuracy and resolution, and estimate the absolute and percentage uncertainty of a reading.
- Round answers to the number of significant figures the data justify, and record means to the precision of the instrument.
- Identify random and systematic errors in a set of results, explain the effect of each on a graph, and suggest a specific improvement for each.
- Organise raw data into a table that scores every mark the on-screen examination gives for it.
You met this before
- A measurement is a number and a unit: 2.5 m, 0.30 kg, 12.7 s. The unit tells the reader what the number counts (Year 1).
- Length is measured with a rule in metres, mass with a balance in kilograms, time with a stopwatch in seconds, volume of a liquid with a measuring cylinder in cm³ — read at eye level (Year 1).
- Repeating a reading three times and taking the mean makes scatter cancel; an anomalous result is one that does not fit the pattern of the others and is left out of the mean (Year 1).
- Tables have the quantity and its unit in the heading only, with the independent variable in the first column and results in order (Year 1).
- Weight is a force: W = mg, with g = 9.8 N/kg on Earth, so a 0.50 kg mass weighs 4.9 N (Year 3).
A student measures a table as 150 cm long and 75 cm wide and its mass as 12 kg. Write the length, width and mass in SI base units, and calculate the table’s weight.
Reveal
1.50 m, 0.75 m and 12 kg (the kilogram is already the base unit).
W = mg
W = 12 kg × 9.8 N/kg
W = 118 N ≈ 120 N (two significant figures)
Notice that 150 cm became 1.50 m, not 1.5 m — the rule read to the nearest centimetre, so the answer keeps three significant figures.
Revise it in full: Year 1 · Being a scientist · Lesson 2 — Measuring: how do you know your number is right? · Year 1 · Being a scientist · Lesson 4 — Tables and graphs: making the pattern visible · Year 3 · Mechanics · Lesson 3 — Gravity everywhere
The puzzle
Two students time a pendulum. Amo times one swing with a stopwatch, five times over, and gets 1.3, 1.5, 1.2, 1.6 and 1.4 s. Kago times ten swings, once, and gets 14.2 s.
Amo has five readings and a mean; Kago has one number. Yet Kago’s value for the period, 1.42 s, is the better one — and a physicist can say by how much.
The stopwatch shows hundredths of a second, so why is neither student entitled to write 1.42 s as if it were exact? And how can a single reading beat a mean of five?
Predict what limits the stopwatch reading: the watch, or the thumb pressing it. Then predict whether timing 20 swings instead of 10 would make Kago’s answer better, worse or the same.
Compare your answer
The thumb. A stopwatch resolves 0.01 s, but the signal from eye to brain to thumb takes about 0.2 s, and that delay varies from press to press — so any single timing is uncertain by roughly ±0.2 s however many digits the display shows. Timing 20 swings shares that same ±0.2 s across twenty periods, so the uncertainty in one period falls to ±0.01 s: better still. Section 1.4 puts numbers on it.
The science — structured notes
1.1 The SI: seven base units and everything built from them
Every physical quantity in this course is measured in the International System of Units (SI). Seven base units are defined; every other unit is derived by multiplying and dividing them.
Since 20 May 2019 all seven have been defined from constants of nature — the kilogram, for example, is fixed by the Planck constant rather than by a metal cylinder in a vault near Paris — so a laboratory anywhere on Earth can realise the same units without shipping an object around.
| Quantity | Base unit | Symbol | Instrument in a school laboratory |
|---|---|---|---|
| Length | metre | m | Metre rule, tape, vernier callipers, micrometer |
| Mass | kilogram | kg | Digital balance (reads in g — convert) |
| Time | second | s | Stopwatch, light gate and datalogger |
| Electric current | ampere | A | Ammeter |
| Temperature | kelvin | K | Thermometer (reads in °C; K = °C + 273) |
| Amount of substance | mole | mol | Calculated from mass (Chemistry) |
| Luminous intensity | candela | cd | Light meter (rarely used in the MYP) |
Derived units you will use in this unit: area m², volume m³, density kg/m³ (also written kg m⁻³).
- force: newton, 1 N = 1 kg m s⁻²
- pressure: pascal, 1 Pa = 1 N/m²
- pressure in base units: 1 Pa = 1 kg m⁻¹ s⁻²
- energy: joule, 1 J = 1 N m
Writing a unit as “N/m²” and as “N m⁻²” means the same thing; the on-screen paper accepts both, but it does not accept “m/s^2” typed with a caret — the May 2021 physics markscheme (Q3(b)) awarded the unit mark for m s⁻² only with a proper superscript, so find the superscript tool before exam day.
| Prefix | Symbol | Factor | Example | Prefix | Symbol | Factor | Example |
|---|---|---|---|---|---|---|---|
| giga | G | 10⁹ | 1 GW power station | centi | c | 10⁻² | 1 cm = 0.01 m |
| mega | M | 10⁶ | 38 MPa at the Titanic wreck | milli | m | 10⁻³ | 1 mm = 0.001 m |
| kilo | k | 10³ | 101 kPa of atmosphere | micro | μ | 10⁻⁶ | 0.55 μm: green light |
| (none) | — | 10⁰ | 1 Pa | nano | n | 10⁻⁹ | 5 nA leakage current |
Converting areas and volumes is the trap. 1 cm = 10⁻² m, so the area and volume factors are the square and the cube of that:
1 cm² = (10⁻²)² m² = 10⁻⁴ m²
1 cm³ = (10⁻²)³ m³ = 10⁻⁶ m³
A 5 cm² piston is 5 × 10⁻⁴ m²; a 250 cm³ measuring cylinder holds 2.5 × 10⁻⁴ m³.
Written in scientific notation, a × 10ⁿ with 1 ≤ a < 10, a number shows its size and its significant figures at a glance: 0.000 020 s is 2.0 × 10⁻⁵ s.
The May 2021 physics paper gave a whole mark for writing an answer in standard form even when the value was wrong.
Convert: (i) 89.9 kPa to Pa; (ii) 12 cm² to m²; (iii) 0.75 g/cm³ to kg/m³; (iv) 3.2 × 10⁻⁵ m to μm.
Reveal answer
(i) 89.9 × 10³ Pa = 8.99 × 10⁴ Pa (89 900 Pa)
(ii) 12 × 10⁻⁴ m² = 1.2 × 10⁻³ m²
(iii) 1 g = 10⁻³ kg and 1 cm³ = 10⁻⁶ m³
so 1 g/cm³ = 10⁻³ ÷ 10⁻⁶ kg/m³ = 1 000 kg/m³
0.75 g/cm³ = 750 kg/m³
(iv) 3.2 × 10⁻⁵ m = 32 × 10⁻⁶ m = 32 μm
1.2 Choosing the instrument: resolution, and how to read it
The resolution of an instrument is the smallest change it can show — the smallest scale division, or the last digit of a display. Choose the instrument whose resolution is small compared with the quantity you are measuring: a metre rule (1 mm) is fine for a 30 cm ramp and useless for a 0.4 mm wire, where you need a micrometer (0.01 mm).
| Instrument | Resolution | Typical uncertainty of one reading | How to read it well |
|---|---|---|---|
| Metre rule | 1 mm | ± 0.5 mm at each end, so ± 1 mm for a length | Start at a mark, not the worn end; eye directly above the mark to avoid parallax |
| Vernier callipers | 0.1 mm (0.02 mm on good ones) | ± 0.1 mm | Close on the object gently; check the zero with the jaws closed |
| Micrometer | 0.01 mm | ± 0.01 mm | Use the ratchet so every reading has the same pressure; subtract the zero error |
| Digital balance | 0.1 g or 0.01 g | ± last digit | Tare (zero) with the empty container; shield from draughts; object dry |
| Measuring cylinder 50 cm³ | 1 cm³ | ± 0.5 cm³ | On a flat bench, eye level with the bottom of the meniscus |
| Burette | 0.1 cm³ | ± 0.05 cm³ | Read the bottom of the meniscus; note the scale runs downwards |
| Stopwatch | 0.01 s | ± 0.2 s — set by the human reaction time, not the display | Time many events (10 swings); count “zero” on release |
| Light gate and datalogger | 0.001 s | ± 0.001 s | Align the beam with the object’s edge; enter the object’s width correctly |
| Thermometer (alcohol) | 1 °C | ± 0.5 °C | Bulb fully in the liquid, not touching the glass; wait for the reading to settle |
Two habits protect every reading. Eye level: read a scale with your eye directly in line with the mark, as the diagram shows — the same rule applies to the top of a ruler mark and the pointer of a newton meter. Zero check: before measuring, see what the instrument reads with nothing on it.
A balance showing 0.3 g empty, or a micrometer showing 0.02 mm closed, has a zero error that must be subtracted from every reading; the November 2021 physics paper accepted “a zero error on the meter” as the source of a systematic error, and Section 1.6 explains why that word matters.
1.3 Precision, accuracy and resolution are three different words
| Word | Meaning | What improves it | What it looks like in results |
|---|---|---|---|
| Resolution | The smallest change the instrument can display | A better instrument (micrometer instead of rule) | More decimal places that mean something |
| Precision | How closely repeated readings agree with each other | Repeats and a mean; a steadier method (light gate, not thumb) | Small range between trials: 12.6, 12.7, 12.6 s |
| Accuracy | How close a result is to the true value | Calibration; removing zero errors; a correct method | Agreement with the accepted value: g = 9.8, not 8.1 N/kg |
The examiners are strict about the words. A student who “repeats to improve accuracy” is wrong: repeating and averaging improves precision (it shrinks the effect of random error), and does nothing for a zero error that shifts every reading the same way. Both the November 2023 physics markscheme (bouncing balls, Q4) and the May 2024 physics report say so in almost those words.
“A balance that reads to 0.001 g is more accurate than one that reads to 0.1 g.”
Reveal answer
False. It has a finer resolution. If it has not been calibrated, or has a zero error of 0.250 g, every reading is precisely wrong. Accuracy is closeness to the true value and depends on calibration and method, not on the number of digits.
1.4 Uncertainty: how much you are allowed to doubt a reading
No measurement is exact. The absolute uncertainty is the “±” you attach to a reading; the percentage uncertainty is that ± as a percentage of the value:
percentage uncertainty = absolute uncertaintymeasured value × 100 %
Rules of thumb that match what markschemes accept:
- a single reading on an analogue scale: ± half the smallest division (a metre rule: ± 0.5 mm);
- a digital display: ± the last digit (a balance: ± 0.1 g);
- a length found as the difference of two readings on a rule: ± 1 mm (both ends are uncertain);
- a stopwatch: ± 0.2 s, because the reaction time, not the display, is the limit.
When readings are repeated, the uncertainty is at least half the range (largest minus smallest) of the trials.
Amo times one swing: 1.3, 1.5, 1.2, 1.6 and 1.4 s. Kago times ten swings once: 14.2 s. Estimate the period and its uncertainty from each method, and compare their percentage uncertainties.
Amo: mean = (1.3 + 1.5 + 1.2 + 1.6 + 1.4) ÷ 5
mean = 1.40 s
range = 1.6 − 1.2 = 0.4 s
uncertainty ≈ half the range = ± 0.2 s (the same as the reaction-time estimate)
period = 1.4 ± 0.2 s
percentage uncertainty = 0.2 ÷ 1.4 × 100 = 14 %
Kago: 14.2 ± 0.2 s for ten swings (one start, one stop: the same ± 0.2 s)
period = 14.2 ÷ 10 = 1.42 s
uncertainty shared out: 0.2 ÷ 10 = ± 0.02 s
period = 1.42 ± 0.02 s
percentage uncertainty = 0.02 ÷ 1.42 × 100 = 1.4 %
Comparison: the same watch and the same thumb, but timing a quantity ten times larger divides the percentage uncertainty by ten. That is why Kago may write 1.42 s (three significant figures) and Amo may not write more than 1.4 s. Timing twenty swings would give ± 0.01 s — but a pendulum’s swings shrink, so there is a limit; three trials of ten swings is the usual compromise.
The thickness of one sheet of paper is wanted. A micrometer (± 0.01 mm) reads 0.10 mm for one sheet and 9.85 mm for a stack of 100 sheets. Calculate the percentage uncertainty of each measurement and the thickness of one sheet from the better method.
Reveal answer
One sheet: 0.01 ÷ 0.10 × 100 = 10 %
Stack: 0.01 ÷ 9.85 × 100 = 0.10 % — a hundred times better
Thickness = 9.85 ÷ 100 = 0.0985 mm (± 0.0001 mm), which is 98.5 μm
The single-sheet reading could only justify “0.10 mm”.
1.5 Significant figures: writing only the digits you have earned
Significant figures (s.f.) are the digits that carry information: all non-zero digits, zeros between them (2.05 has three), and trailing zeros after a decimal point (2.50 has three — the zero says “measured to the hundredth”). Leading zeros are not significant (0.0025 has two). A raw reading is written to the instrument’s resolution, no more and no less: a rule gives 0.250 m, not 0.25 m; a balance gives 54.0 g, not 54 g.
When you calculate, the answer cannot be more certain than the least certain number that went into it: round the answer to the smallest number of significant figures in the data (for multiplying and dividing), or as the question instructs.
Keep full precision on the calculator until the final line, then round once. The markschemes say exactly this — “43 Ω to two significant figures, the rounding carrying the third mark” (Integrated Sciences, May 2022, Q3(b)); “4.80 correctly rounded to 2 d.p. to match the table” (Physics, May 2021, Q4(e)).
A student wants the density of a coin. Its mass on a 0.1 g balance is 7.5 g. Its volume, from the rise of water in a 50 cm³ measuring cylinder (± 0.5 cm³), is 0.9 cm³. (1) Calculate the density. (2) Find the percentage uncertainty of the mass and of the volume, and state which dominates. (3) Decide how many significant figures the density deserves. (4) Suggest the improvement that would fix the problem.
(1) ρ = m ÷ V
ρ = 7.5 g ÷ 0.9 cm³
ρ = 8.333… g/cm³
(2) Mass: 0.1 ÷ 7.5 × 100 = 1.3 %. Volume: 0.5 ÷ 0.9 × 100 = 56 %. The volume dominates completely — the measuring cylinder’s resolution is nearly as big as the coin.
(3) A 56 % uncertainty means the density could be anywhere from about 5 to 19 g/cm³. Writing 8.333 g/cm³ claims a precision that does not exist; even 8.3 is generous. Honest answer: ρ ≈ 8 g/cm³ (one significant figure), consistent with a copper alloy such as cupronickel (8.9 g/cm³) but not able to prove it.
(4) Measure twenty identical coins together: mass 150.0 g (± 0.1 g, 0.07 %), volume 17.0 cm³ (± 0.5 cm³, 2.9 %).
ρ = m ÷ V
ρ = 150.0 g ÷ 17.0 cm³
ρ = 8.82 g/cm³, with about 3 % uncertainty
Three significant figures are now earned, and the value (8.8 ± 0.3 g/cm³) is consistent with cupronickel. Multiplying the quantity, not the number of repeats, is the fix.
State the number of significant figures in each: 0.0450 kg; 1 200 m; 3.00 × 10⁸ m/s; 101.3 kPa. Then calculate 7.85 g ÷ 2.9 cm³ and give the answer to the correct number of significant figures.
Reveal answer
0.0450 kg — 3 (the trailing zero counts; the leading zeros do not). 1 200 m — ambiguous, 2, 3 or 4; write 1.20 × 10³ m to make it 3. 3.00 × 10⁸ — 3. 101.3 — 4. 7.85 ÷ 2.9 = 2.7068… → 2.7 g/cm³, two significant figures, because 2.9 has only two.
1.6 Random and systematic error: two enemies, two weapons
| Random error | Systematic error | |
|---|---|---|
| What it is | Unpredictable variation from reading to reading — sometimes high, sometimes low | A shift in the same direction by the same amount (or the same fraction) in every reading |
| Typical causes | Reaction time on a stopwatch; judging a moving ball against a rule; draughts on a balance; a fluctuating supply; estimating between scale marks | A zero error (balance reading 0.3 g empty; micrometer not closing to 0); a mis-calibrated instrument; a stopwatch always started late; a rule read from the same angle every time (parallax); an instrument that warms up during the experiment; measuring to the wrong point (the bottom of a ball instead of its centre) |
| Effect on results | Scatter: points spread either side of the line; the mean is still about right | Every point displaced the same way: the line keeps its gradient but is shifted, so it misses the origin (a false intercept); the mean is wrong |
| How to reduce it | Repeat and average; use an instrument that removes the judgement (light gate, datalogger); measure a larger quantity | Check and subtract zero errors; calibrate against a known standard; change the method; repeating does nothing |
| Test for it | Trials disagree with each other | Trials agree with each other but the graph misses the origin, or the result disagrees with the accepted value |
Notice what the timeline says: a late start and a late stop partly cancel, so the stopwatch is not always 0.2 s out — but the two delays are never exactly equal, and they change from trial to trial. That variation is what makes reaction time a random error.
If the student always starts on the word “go” from a partner who says it slightly before releasing, the start is consistently early: that is systematic. The answer examiners want is never a vague “the person made a mistake” — the May 2024 and May 2025 physics reports both refused such answers — but the specific error, named, with its direction and its effect.
The right-hand graph is the November 2021 physics paper’s own case (Q5): resistance against length for a wire should pass through the origin, because a wire of zero length has zero resistance; the line had an intercept of about 0.008 Ω, which is evidence of a systematic error — the leads and contacts adding the same resistance to every reading.
And the markscheme’s second point is the one students miss: because every value is shifted by the same amount, the offsets cancel in Δy ÷ Δx, so the gradient is still accurate.
A systematic error ruins an intercept and a single value; it leaves a gradient alone.
For each, say random or systematic, give its effect, and name one improvement: (i) a balance reads 0.3 g with nothing on the pan; (ii) a student reads a measuring cylinder standing up, looking down at it, for every reading; (iii) the room’s air conditioner makes the balance flicker between 12.4 and 12.6 g; (iv) a thermometer’s bulb touches the hot base of the beaker.
Reveal answer
(i) Systematic zero error — every mass 0.3 g too high; tare the balance or subtract 0.3 g. (ii) Systematic parallax — every volume read too high (Section 1.2); read at eye level with the bottom of the meniscus.
(iii) Random — readings scatter by ± 0.1 g; shield the balance with a box or switch the air off, and take the mean of three. (iv) Systematic — every temperature too high, by an amount that grows as the beaker heats; hold the bulb in the middle of the liquid with a clamp.
A class measures g by timing a ball’s fall. Every group’s value is between 8.0 and 8.4 N/kg, agreeing closely, but the accepted value is 9.8 N/kg. Explain what the agreement and the disagreement each tell you, and suggest the cause.
Reveal answer
Close agreement between groups means small random error: the method is precise. Disagreement with 9.8 by a consistent 15 % or so means a systematic error shifting every result the same way — the results are not accurate.
Since g = 2h ÷ t², a value that is too low means t is measured too large: the stopwatch is being stopped late (the student waits to hear the ball land, and the sound and the thumb both add delay) more than it is started late.
Improvement: a light gate at the bottom, or an electromagnet release that starts the timer.
1.7 Tables and means: the marks the examiners give for tidiness
“Organize and present the data in a table” carries several marks on the on-screen paper (six in November 2023 Physics Q5(c), including the calculated value), and the same four features earn them every session:
- the quantity and its unit in the heading only — “Length / m”, never a unit in each cell;
- the independent variable in the first column, rows in ascending order;
- every value in a column to the same number of decimal places, matching the instrument;
- values converted to one unit per column (the November 2023 markscheme docked a mark for 50.0 cm sitting in a column of metres).
Then the mean: calculated leaving out any anomaly, and rounded to the precision of the raw readings — a mean of 12.6, 12.7 and 12.6 s is 12.6 s, not 12.63 s.
A notebook page reads: “length 40cm — 12.7s, 12.6 s, 12.61s; 20 cm — 8.9, 9.0, 9.1 secs; 0.6 m — 15.5 s, 15.6 s, 15.4 s; mean for 40 cm = 12.637”. Identify five things that must change before this becomes an exam-standard table.
Reveal answer
(1) Units into the headings only: “Length / m” and “Time for 10 swings / s”. (2) One unit per column: 0.20, 0.40 and 0.60 m (or 20, 40, 60 cm) — not a mixture.
(3) Rows in ascending order of the independent variable: 0.20, 0.40, 0.60 m. (4) Consistent decimals matching the stopwatch, 1 d.p.: 12.61 becomes 12.6, and 8.9 stays 8.9 (the stopwatch here is being read to 0.1 s).
(5) The mean rounded to the instrument: 12.6 s, not 12.637 s. A sixth: a heading “Trial 1 / Trial 2 / Trial 3 / Mean” so the reader knows which is which.
The page rewritten as an exam-standard table:
| length / m | time for 10 swings trial 1 / s | trial 2 / s | trial 3 / s | mean time for 10 swings / s |
|---|---|---|---|---|
| 0.20 | 8.9 | 9.0 | 9.1 | 9.0 |
| 0.40 | 12.7 | 12.6 | 12.6 | 12.6 |
| 0.60 | 15.5 | 15.6 | 15.4 | 15.5 |
A student’s pendulum results for T² against L give a straight line that cuts the T² axis at +0.05 s² instead of passing through the origin. What does this tell you, what is the most likely cause, and is the value of g from the gradient still trustworthy?
Reveal answer
A pendulum of zero length would have zero period, so the line should pass through the origin: the intercept is evidence of a systematic error. Most likely the length was measured to the top of the bob rather than its centre, so every length is short by the bob’s radius — every T² is “too big for its L” by the same amount.
The gradient (4π² ÷ g) is unaffected, because the shift is the same for every point, so g from the gradient is still trustworthy — exactly the argument of the November 2021 physics markscheme.
A density is calculated as m ÷ V with m = 54.1 g (± 0.1 g) and V = 20 cm³ (± 0.5 cm³). Show that the percentage uncertainties add when quantities are divided, find the absolute uncertainty in the density, and decide whether the result distinguishes aluminium (2.70 g/cm³) from glass (2.50 g/cm³).
Reveal answer
Largest possible ρ = 54.2 ÷ 19.5 = 2.779 g/cm³
Smallest possible ρ = 54.0 ÷ 20.5 = 2.634 g/cm³
Central value ρ = 54.1 ÷ 20 = 2.705 g/cm³
Half the spread = ± 0.07 g/cm³, which is 2.7 % of 2.705
Check: 0.18 % (mass) + 2.5 % (volume) = 2.7 % — the percentage uncertainties add
So ρ = 2.71 ± 0.07 g/cm³: the range 2.64–2.78 includes aluminium and excludes glass (2.50) — the measurement does distinguish them, but only just; a 5 cm³ sample (± 11 %, 2.41–2.99 g/cm³) would not.
This is why Lesson 2’s lab uses a gradient of mass against volume rather than one reading.
Try it — the pendulum as a measurement laboratory
The pendulum lab times ten swings of a bob on a string. You set the length from 10 to 100 cm, and you can also change the bob’s mass and the release angle — two variables that, as you will find, the period does not care about. Every timing carries a realistic random error of about 1.5 %, so three trials will not agree exactly; that scatter is the subject of this lesson.
Your three trials at 0.60 m were 15.5, 15.6 and 15.4 s. Calculate the period with its uncertainty and percentage uncertainty, and explain why the lab’s own model value of 15.54 s should not be written to four significant figures in your table.
Reveal answer
Mean = 15.5 s (1 d.p., matching the readings); range 0.2 s, so ± 0.1 s from the spread — but the reaction-time estimate of ± 0.2 s is larger, so use ± 0.2 s.
T = 1.55 ± 0.02 s
percentage uncertainty = 0.02 ÷ 1.55 × 100 = 1.3 %
A four-figure 15.54 s would claim ± 0.005 s, forty times better than a thumb can deliver; the stopwatch’s display resolution is not the measurement’s uncertainty.
Common mistakes to avoid
- “The person made a mistake.” Never credited. Name the error: reaction time on the stopwatch (random), parallax from reading above the meniscus (systematic), a zero error on the balance (systematic).
- “Repeat to improve accuracy.” Repeating improves precision by averaging out random error; it cannot remove a systematic shift.
- Confusing resolution with uncertainty. A stopwatch shows 0.01 s; the reading is uncertain by 0.2 s.
- Too many figures. 12.637 s from readings to 0.1 s; 8.333 g/cm³ from a 56 % uncertain volume. Round once, at the end, to what the data justify.
- Too few figures. Writing 20 for a balance reading of 20.0 g throws away the evidence of the resolution (May 2025 physics report).
- Areas converted like lengths. 5 cm² is 5 × 10⁻⁴ m², not 5 × 10⁻² m².
- Units in every cell, mixed units in one column, rows out of order. The four table marks are the easiest on the paper and the most often dropped.
- Calling a lack of repeats an error. “A lack of repeated trials is not a source of error” (May 2024 physics report).
Study smart: how this is examined
Study routine. Today: write the seven base units and the prefix ladder on one card, and the random/systematic table on another; do the coin EXAMPLE with the book closed. In two days: build an exam-standard table from the notebook page in the Spot-the-errors check, then time ten swings of any pendulum you can rig (a key on a shoelace) three times and write T ± ΔT.
In a week: run the lab as a full investigation, draw T² against L, get g from the gradient, and take the quiz.
How we found out — from a king’s foot to a constant of nature
For most of history a unit was a thing: a bronze bar, a royal foot, a grain of barley. Trade between towns with different bars was a permanent quarrel, and science could not compare results across borders — the problem the May 2021 physics paper asked about with Galileo’s water clock, which measured time in “volumes of water collected” that no other laboratory could reproduce.
In 1791, in the middle of the French Revolution, the Academy of Sciences proposed a unit taken from the Earth itself: the metre as one ten-millionth of the distance from the North Pole to the equator through Paris.
Jean-Baptiste Delambre and Pierre Méchain spent six years (1792–1798) surveying the meridian from Dunkirk to Barcelona to fix it. In 1799 a platinum metre and a platinum kilogram (the mass of a cubic decimetre of water) were deposited in the Archives.
Méchain’s survey, it later turned out, carried a small systematic error; the metre was about 0.2 mm short of its definition — and nobody could change it, because by then the bar was the metre.
The Metre Convention of 1875, signed by seventeen countries, created the International Bureau of Weights and Measures (BIPM) at Sèvres, and in 1889 new platinum–iridium standards — the International Prototype Metre and the International Prototype Kilogram (IPK) — were distributed as copies to member states.
The system was named the SI at the 11th General Conference on Weights and Measures in 1960, when the metre was redefined by a wavelength of krypton light; in 1983 it became the distance light travels in 1/299 792 458 of a second, tying length to the speed of light.
The kilogram held out longest: comparisons showed the IPK drifting by tens of micrograms relative to its copies, and a cylinder that defines mass cannot itself be checked. On 20 May 2019 all seven base units were redefined from fixed constants — the kilogram from the Planck constant, realised with a Kibble balance that weighs a mass against an electromagnetic force.
The unit of mass is now a number of nature, the same in Gaborone as in Sèvres, and no vault is needed.
The IPK was found to differ from its official copies by about 50 μg over a century. Explain why this was a problem that repeating the comparison could never solve, and what kind of error it represents.
Reveal answer
By definition the IPK’s mass was exactly 1 kg, so if it lost or gained atoms (surface contamination, cleaning), every mass in the world defined against it shifted by the same fraction — a systematic error in the standard itself.
Repeating comparisons only measured the drift more precisely; it could not say whether the IPK or the copies had changed, because nothing more fundamental existed to check against. Defining the kilogram from the Planck constant removes the artefact: any laboratory can realise the unit and check another, so the standard can no longer drift.
Where it is used now
- Shops and fuel pumps. In every country, weights-and-measures (legal metrology) inspectors check the balances in butcheries and the meters on fuel pumps against reference masses and volumes traceable to the SI, so that “1 kg” in one shop is 1 kg in the next. A zero error on a shop scale is a systematic error that costs customers money on every purchase.
- Medicine. A dose of 5 mg and one of 5 μg differ by a thousand; a blood-glucose meter’s reading is quoted with its uncertainty; a baby’s mass on a clinic balance is recorded to 0.01 kg because a 20 g change matters.
- Building. A surveyor’s total station measures to 1 mm over 100 m; a builder’s tape to 1 mm over 5 m; a bridge that is 30 mm out over 300 m will not meet in the middle.
- Sport. Athletics timing to 0.01 s with a fully automatic system started by the gun’s sound and stopped by a photo-finish camera — a light gate for humans; hand timing is not accepted for records, for the reason in Section 1.6.
- Satellites. GPS positions your phone to a few metres by timing radio signals to nanoseconds; a clock error of 1 μs is a position error of 300 m, so the satellites carry atomic clocks and are corrected for relativity.
- Mining and agriculture. Ore grades quoted in grams per tonne (parts per million), rainfall to 0.1 mm, borehole water levels to a centimetre — every one a measurement with a resolution, an uncertainty and a possible zero error.
Science now — and the scientist you need to be
What is happening in 2026
The 2019 SI is still being pushed to its limits. National laboratories are building Kibble balances small enough for industry, so that a factory could one day realise the kilogram on its own bench rather than sending a mass away for calibration.
Optical atomic clocks — which count the oscillations of a strontium or ytterbium atom at hundreds of trillions of hertz — now outperform the caesium clocks that define the second by a factor of a hundred, and the international committees have set out a road map towards redefining the second from an optical transition around the end of this decade.
Measurement is also going wherever the data are: a modern smartphone carries a barometer, an accelerometer, a magnetometer and a temperature sensor, each a micro-machined chip with a resolution, a zero error and a drift that its software corrects against known references.
Machine-learning models are increasingly used to calibrate sensors — learning a sensor’s systematic error against a reference and subtracting it — and to flag anomalous readings in networks of thousands of weather stations, water meters and medical devices.
None of that changes the discipline of this lesson; it makes it more valuable, because a model trained on data with an uncorrected zero error learns the error too. Careers that live on it: metrology, instrumentation engineering, quality control, surveying, and every laboratory science.
The mindshift
What you need to be able to do
Read any instrument correctly and say its resolution and its uncertainty. Convert units, including areas and volumes, without a slip. Write a raw-data table that scores all four table marks and a mean that matches the instrument. And when results go wrong, say which error, in which direction, with what effect — and fix it with a change that a laboratory could actually make.
See an example
Zero error 2 g; masses 203, 199, 201, 204, 198 g → mean 201 g, range 6 g, ± 3 g (1.5 %); jug ± 10 cm³ (5 %) dominates; ρ = 1.0 ± 0.07 g/cm³. The assistant gave good definitions but claimed “repeating improves accuracy and precision” until asked to distinguish random from systematic error.
Key terms
| SI base unit | One of the seven defined units (m, kg, s, A, K, mol, cd) from which all other units are derived. |
| Derived unit | A unit built from base units by multiplying and dividing: N = kg m s⁻², Pa = N m⁻², kg/m³. |
| Prefix | A multiplier attached to a unit: k = 10³, m = 10⁻³, μ = 10⁻⁶, n = 10⁻⁹, M = 10⁶, G = 10⁹. |
| Resolution | The smallest change an instrument can display: the smallest scale division or the last digit. |
| Precision | How closely repeated readings agree with one another; improved by averaging and steadier methods. |
| Accuracy | How close a result is to the true value; improved by calibration and removing systematic error. |
| Uncertainty | The ± attached to a reading; absolute (in the unit) or percentage (absolute ÷ value × 100 %). |
| Significant figures | The digits of a value that carry information; an answer is rounded to the fewest s.f. in the data. |
| Random error | Unpredictable variation between readings, causing scatter; reduced by repeating and averaging. |
| Systematic error | A constant shift of every reading in the same direction (zero error, calibration, parallax); not reduced by repeating. |
| Zero error | A non-zero reading with nothing being measured; subtract it from every reading. |
| Parallax | A reading error caused by viewing a scale from an angle; avoided by reading at eye level. |
| Anomaly | A result that does not fit the pattern of the others; excluded from the mean and its cause suggested. |
Think like a scientist
Criterion C A group measures g by dropping a steel ball through a light gate placed just above the floor. They release the ball from rest at a measured height h above the gate, and the datalogger records the fall time t to 0.001 s.
Before starting they wrote: “If the height is increased, the fall time will increase, but t² will be directly proportional to h, because for constant acceleration from rest h = ½gt², so t² = 2h ÷ g.” Results:
| Height h / m | Time trial 1 / s | Time trial 2 / s | Time trial 3 / s | Mean time / s | t² / s² |
|---|---|---|---|---|---|
| 0.50 | 0.325 | 0.327 | 0.326 | ||
| 0.75 | 0.397 | 0.395 | 0.396 | ||
| 1.00 | 0.457 | 0.455 | 0.456 | ||
| 1.25 | 0.508 | 0.510 | 0.552 | ||
| 1.50 | 0.556 | 0.558 | 0.557 |
(a) Identify the anomalous reading and suggest a specific cause. [2]
(b) Complete the mean and t² columns, rounding appropriately, and state the resolution of the timing. [3]
(c) Describe the graph you would draw to test the hypothesis: axes, scales, points and line. [3]
(d) The best-fit line of t² against h has gradient 0.204 s²/m and intercept +0.004 s². Calculate g from the gradient, with its unit. [3]
(e) Explain what the intercept tells you, identify the type of error, suggest its cause, and state whether the value of g in (d) is affected. [4]
(f) State whether the hypothesis is supported, referring to the data. [2]
(g) Compare the random error of this method with timing the fall by stopwatch, using numbers, and state the effect on the graph. [3]
Model answer
(a) Trial 3 at 1.25 m, 0.552 s — 0.04 s above the other two (0.508, 0.510) and almost equal to the time for the greater height 1.50 m, breaking the trend [1]. Cause: the ball clung briefly to the electromagnet (residual magnetism) after the timer had started, or brushed the release clamp and did not fall freely from rest — a specific mechanism, not “a mistake” [1].
(b) Means (anomaly excluded at 1.25 m): 0.326, 0.396, 0.456, 0.509, 0.557 s — 3 d.p., matching the datalogger’s 0.001 s resolution [1 for excluding the anomaly, 1 for the rest]. t²: 0.106, 0.157, 0.208, 0.259, 0.310 s² — 3 d.p. (or 3 s.f.) consistently, unit s² [1].
(c) h (m) on the horizontal axis as the independent variable, t² (s²) on the vertical [1]; axes labelled with quantity and unit, even scales from zero (e.g. 0.25 m and 0.05 s² per large square) filling more than half the grid [1]; five crosses and a ruler line of best fit — not dot-to-dot — which is expected to pass through the origin if t² ∝ h [1].
(d) Gradient = 2 ÷ g, so g = 2 ÷ gradient [1]
g = 2 ÷ 0.204 s²/m [1]
g = 9.8 m s⁻² (or N/kg), with the unit [1]. (Unit: s²/m inverted is m/s².)
(e) If t² ∝ h the line should pass through the origin — zero height, zero time — so a positive intercept means every t² is slightly larger than the height accounts for, or every h is measured slightly too small [1]; this is a systematic error (same shift in every reading), not random [1].
A likely cause: the height was measured to the bottom of the ball but the gate triggers on the ball’s leading edge slightly lower, or the release point was 2 cm above the measured mark — 0.004 s² ÷ 0.204 s²/m = 0.02 m of extra fall [1].
The value of g is unaffected, because a constant offset shifts every point equally and cancels in the gradient Δy ÷ Δx [1].
(f) Supported [1]: t² rose from 0.106 s² at 0.50 m to 0.310 s² at 1.50 m — a factor of 2.9 for a factor of 3.0 in h — and the points lie on a straight line whose gradient gives g = 9.8 m s⁻², close to the accepted 9.81; the small intercept is a systematic offset, not a failure of proportionality [1].
(g) The light gate’s trials agree to ± 0.001 s (0.325–0.327 s), about 0.3 % of a 0.33 s fall; a stopwatch would add a reaction-time uncertainty of about ± 0.2 s, 60 % of the same fall time [1]; the light gate’s random error is therefore some 200 times smaller [1].
On the graph, stopwatch points would scatter widely about the line and t² would be dominated by the timing error (a 0.2 s error in 0.33 s can more than double t²). The light-gate points sit almost exactly on the line — which is why the method allows the small systematic intercept to be seen at all [1].
Probe deeper — exam-style questions
Attempt every question before opening the answer. Marks in brackets show what a full answer is worth.
Label the SI unit of each quantity in p = F ÷ A, and express the unit of pressure in base units.
Reveal answer
F in newtons (N), A in square metres (m²), p in pascals (Pa) [1].
1 Pa = 1 N/m²
1 N = 1 kg m s⁻², so 1 Pa = 1 kg m s⁻² ÷ m²
1 Pa = 1 kg m⁻¹ s⁻² [1]
A radio signal takes 0.000 020 s to cross a bay. Write this time in scientific notation and in microseconds, and calculate the width of the bay if the signal travels at 3.0 × 10⁸ m/s, giving the answer to two significant figures with a unit.
Reveal answer
2.0 × 10⁻⁵ s = 20 μs [1]
distance = speed × time
distance = 3.0 × 10⁸ m/s × 2.0 × 10⁻⁵ s [1]
distance = 6.0 × 10³ m = 6.0 km [1]
(This reverses May 2021 Physics Q1(c), where 6 km gave 2 × 10⁻⁵ s.)
Distinguish between precision and accuracy, and describe one way of improving each in a measurement of the mass of a liquid in a beaker.
Reveal answer
Precision: how closely repeated readings agree; accuracy: how close the result is to the true value [1]; precision — shield the balance from draughts and average three readings [1]; accuracy — tare the balance with the empty beaker (or subtract the beaker’s mass) to remove the systematic error of including the beaker [1].
A student measures the diameter of a wire as 0.42 mm with a micrometer that reads 0.02 mm when closed. Calculate the true diameter, the percentage uncertainty of the corrected reading (± 0.01 mm), and the cross-sectional area in m² to two significant figures.
Reveal answer
Corrected diameter = 0.42 − 0.02 = 0.40 mm [1]
percentage uncertainty = 0.01 ÷ 0.40 × 100 = 2.5 % [1]
area = π(d ÷ 2)²
area = π × (0.20 × 10⁻³ m)²
area = 1.3 × 10⁻⁷ m² [1]
(Forgetting the zero error gives 1.4 × 10⁻⁷ m², a 10 % error in the area from a 5 % error in the diameter — squaring doubles the percentage.)
Students measuring the period of a pendulum obtain a graph of T² against L that is a straight line through the origin, but the gradient gives g = 8.9 N/kg. Identify the kind of error, suggest a cause that affects the gradient and not the intercept, and explain why “repeat the readings” would not help.
Reveal answer
Systematic — every T² is too large by the same fraction (the line is steeper but still through the origin), so the error scales with the reading rather than adding a constant [1].
Cause: an error proportional to the reading, such as a stopwatch that runs about 5 % fast (every T² about 10 % too large), or a rule whose scale makes every length read about 10 % short [1].
Repeating and averaging reduces only random scatter; a proportional calibration error is present in every repeat, so the mean carries the same error — only calibrating the stopwatch or rule against a standard (or changing the instrument) removes it [1].
Organise these notebook readings of a ball’s bounce height into an exam-standard table and calculate the missing mean: drop 1.00 m — bounces 0.62 m, 0.60 m, 0.61 m; drop 50.0 cm — 0.31 m, 0.32 m, 0.30 m; drop 1.50 m — 0.89 m, 0.91 m, 0.90 m; drop 200 cm — 1.17, 1.18, 1.19 m.
Reveal answer
| drop height / m | bounce height trial 1 / m | trial 2 / m | trial 3 / m | mean bounce height / m |
|---|---|---|---|---|
| 0.50 | 0.31 | 0.32 | 0.30 | 0.31 |
| 1.00 | 0.62 | 0.60 | 0.61 | 0.61 |
| 1.50 | 0.89 | 0.91 | 0.90 | 0.90 |
| 2.00 | 1.17 | 1.18 | 1.19 | 1.18 |
Columns: Drop height / m; Bounce height trial 1, 2, 3 / m; Mean bounce height / m — units in the headings only [1]; rows in order 0.50, 1.00, 1.50, 2.00 m with every height converted to metres and 2 d.p.
(50.0 cm → 0.50 m; 200 cm → 2.00 m) [1]; consistent 2 d.p. in every column [1]; means 0.31, 0.61, 0.90, 1.18 m — the 2.00 m mean is (1.17 + 1.18 + 1.19) ÷ 3 = 1.18 m, rounded to the 0.01 m of the readings [1].
(This is the format of November 2023 Physics Q5(c), where a “many errors” table had to be rebuilt.)
7. A cylinder’s volume is found from its diameter d = 2.00 cm (± 0.01 cm, callipers) and height h = 10.0 cm (± 0.1 cm, rule) using the formula below. Calculate V, the percentage uncertainty in V (percentage uncertainties add, and a squared quantity counts twice), and the absolute uncertainty. Then decide whether measuring the height with callipers (± 0.01 cm) would be worth the trouble.
V = π(d ÷ 2)²h
Reveal answer
V = π × (1.00 cm)² × 10.0 cm
V = 31.4 cm³ [1]
% uncertainty in d = 0.01 ÷ 2.00 × 100 = 0.5 %
% uncertainty in h = 0.1 ÷ 10.0 × 100 = 1.0 %
% uncertainty in V = 2 × 0.5 % + 1.0 % = 2.0 % [1]
absolute uncertainty = 0.02 × 31.4 = ± 0.6 cm³
V = 31.4 ± 0.6 cm³ [1]
With callipers for the height the height term falls to 0.1 %, total 1.1 % (± 0.35 cm³) — the diameter term now dominates because it is squared. The improvement is worth having, but the next step would be a micrometer on the diameter [1].
8. A pressure sensor lowered into a tank gives absolute pressures of 101, 111, 121, 130, 140 and 150 kPa at depths 0, 1, 2, 3, 4 and 5 m.
A second sensor, wrongly zeroed, gives 95, 105, 115, 124, 134 and 144 kPa. Sketch (describe) both lines, state what each sensor gives for the gradient and for the pressure at the surface, and explain which of the two results from the faulty sensor can still be trusted and why.
Reveal answer
Both are straight lines of the same gradient, about 9.8 kPa/m; the second lies 6 kPa below the first at every depth [1]; the good sensor gives an intercept of 101 kPa (atmospheric pressure) and the faulty one 95 kPa [1].
The faulty sensor’s gradient can be trusted, because a zero error is a constant offset that cancels in Δp ÷ Δh [1]. So the density is still found from it:
ρ = gradient ÷ g
ρ = 9 800 Pa/m ÷ 9.8 N/kg
ρ = 1 000 kg/m³
Its intercept cannot be trusted, because the offset sits entirely in the intercept, so its “atmospheric pressure” is 6 kPa low [1].
(Lesson 4 makes p = patm + ρgh the subject of a whole lab.)
Summary
- The SI has seven base units (m, kg, s, A, K, mol, cd); everything else is derived. Prefixes are powers of ten; areas and volumes convert with the power squared or cubed (1 cm³ = 10⁻⁶ m³; 1 g/cm³ = 1 000 kg/m³).
- Resolution is what the instrument can show; precision is how well repeats agree; accuracy is closeness to the truth. Read at eye level, to the bottom of the meniscus, and subtract any zero error.
- Uncertainty: ± half a division (analogue), ± the last digit (digital), ± 0.2 s for a stopwatch; percentage uncertainty = absolute ÷ value × 100 %. Measure a larger quantity (10 swings, 100 sheets) to shrink it.
- Write raw readings to the instrument’s resolution and round calculated answers once, to the fewest significant figures in the data or as instructed.
- Random error scatters points and is reduced by averaging; systematic error shifts every point the same way, creates a false intercept, leaves the gradient intact, and is removed only by calibration or a better method. Never blame a vague “mistake”: name the error.
- Tables: units in the headings only, independent variable first, rows in order, consistent decimals, one unit per column; means exclude anomalies and match the instrument.
Next time → Density: mass per unit volume, how to measure it for blocks, stones, liquids and gases, the particle picture of why lead is heavy and why ice floats — and a lab where the gradient is the answer.