Significant figures and units in IB Physics: the rules, with examples
By Evan Kotronis · Updated 19 October 2026
In short
Give a calculated answer to the same number of significant figures as the least precise value in the data, or one more. Keep the full calculator value until the last line and round once. Every final answer needs a unit, and comparing the units on both sides of an equation is the quickest way to catch a wrong formula.
In one minute
7 points
- Significant figures show how precisely a value is known: 2.5 m and 2.50 m are not the same measurement.
- Leading zeros never count (0.0042 has 2 s.f.); trailing zeros after a decimal point always count (4.200 has 4 s.f.).
- For multiplying and dividing, match the least precise data value; for adding and subtracting, match the fewest decimal places.
- Round only the final answer. Rounding in the middle of a calculation can change the third significant figure.
- Every final answer needs a unit. The SI base units are kg, m, s, A, K and mol; every other unit can be written in terms of them.
- If the units on the two sides of an equation do not match, the equation is wrong. If they do match, it may still be missing a number such as ½.
- Quote an uncertainty to one significant figure and round the value to the same decimal place.
Contents
13 sections
- What do significant figures tell you?
- How do you count significant figures?
- How many significant figures should an answer have?
- When should you round?
- How do standard form and prefixes work?
- What are the SI base units?
- How do you check an equation with units?
- How should units be written?
- How do significant figures and uncertainties fit together?
- Exam tips
- Common misconceptions
- Key rules
- Check your understanding
What do significant figures tell you?
A measured value is never exact. The number of significant figures (s.f.) tells the reader how precisely it was measured. A length written as 2.5 m means "somewhere close to 2.5 m, to the nearest 0.1 m". Written as 2.50 m, it means the length was measured to the nearest centimetre. The extra zero carries information, so it must not be dropped or added carelessly.
The same idea applies to a calculated answer. A calculator shows ten digits, but the answer cannot be more precise than the measurements that went into it. Writing 4.0000 m s−1 from data measured to two significant figures claims a precision that nobody has.
How do you count significant figures?
Start at the first non-zero digit and count every digit after it, with one exception: trailing zeros in a whole number with no decimal point are ambiguous.
| Value | Significant figures | Why |
|---|---|---|
| 3.07 | 3 | A zero between non-zero digits counts. |
| 0.0042 | 2 | Leading zeros only place the decimal point; they never count. |
| 4.200 | 4 | Trailing zeros after a decimal point were measured, so they count. |
| 100.0 | 4 | The decimal point shows that every zero was measured. |
| 2500 | 2, 3 or 4 | Ambiguous. Write 2.5 × 103 (2 s.f.) or 2.50 × 103 (3 s.f.) to make it clear. |
| 6.02 × 1023 | 3 | In standard form, count the digits in the number before the power of ten. |
Standard form removes every ambiguity, which is why it is the safest way to write very large or very small answers.
How many significant figures should an answer have?
Multiplying and dividing. Give the answer to the same number of significant figures as the least precise value in the data. One more is also acceptable as a rule of thumb; many more is not.
A trolley travels 12.4 m in 3.1 s. Find its average speed.
\(v = \dfrac{d}{t} = \dfrac{12.4}{3.1} = 4.0000\ldots\)
The time has 2 s.f., the distance has 3. The least precise value has 2 s.f., so the answer is 4.0 m s−1. Writing "4" would lose the zero that shows the precision; writing "4.000" would claim too much.
Adding and subtracting. Here the decimal places matter, not the significant figures. The answer cannot have more decimal places than the value with the fewest.
Two lengths, 12.4 cm and 0.356 cm, are added.
12.4 + 0.356 = 12.756. The first length is known only to one decimal place, so the sum is 12.8 cm.
Subtraction can lose precision fast: 36.4 °C − 21.9 °C = 14.5 °C keeps three significant figures, but 22.4 °C − 21.9 °C = 0.5 °C has only one.
Exact numbers do not limit the answer. Counted quantities (10 oscillations, 3 resistors), the ½ in \(\tfrac12mv^2\), the 2 in \(2\pi\) and the value of π itself are exact. Only measured values and given data set the precision.
When should you round?
Round once, at the end. Keep the full calculator value for every intermediate step, either in the calculator's memory or by writing down at least two more figures than you need.
A force of 125 N acts on a circular piston of radius 0.0734 m. Find the pressure.
Area: \(A = \pi r^2 = \pi \times 0.0734^2 = 0.016925\ldots\ \text{m}^2\)
Pressure: \(p = \dfrac{F}{A} = \dfrac{125}{0.016925\ldots} = 7385.3\ldots\ \text{Pa}\)
The data have 3 s.f., so the answer is 7.39 × 103 Pa.
If the area is rounded to 0.017 m2 first, the pressure comes out as 7353 Pa, or 7.35 × 103 Pa: wrong in the third significant figure, even though every step looked reasonable.
How do standard form and prefixes work?
Standard form writes a number as \(a \times 10^n\) with \(1 \le a < 10\). Prefixes are shorthand for powers of ten and must be converted before a value goes into an equation.
| Prefix | Symbol | Factor | Example |
|---|---|---|---|
| pico | p | 10−12 | 4.7 pF = 4.7 × 10−12 F |
| nano | n | 10−9 | 589 nm = 5.89 × 10−7 m |
| micro | μ | 10−6 | 25 μA = 2.5 × 10−5 A |
| milli | m | 10−3 | 0.25 mm = 2.5 × 10−4 m |
| kilo | k | 103 | 4.7 kΩ = 4.7 × 103 Ω |
| mega | M | 106 | 1.6 MeV = 1.6 × 106 eV |
| giga | G | 109 | 2.4 GHz = 2.4 × 109 Hz |
Watch squared and cubed units. One centimetre is 10−2 m, so one square centimetre is (10−2 m)2 = 10−4 m2, and one cubic centimetre is 10−6 m3.
What are the SI base units?
Every quantity in the course can be expressed in six base units. Derived units such as the newton or the volt are combinations of them.
| Quantity | Base unit | Symbol |
|---|---|---|
| Mass | kilogram | kg |
| Length | metre | m |
| Time | second | s |
| Electric current | ampere | A |
| Temperature | kelvin | K |
| Amount of substance | mole | mol |
| Derived unit | Comes from | In base units |
|---|---|---|
| newton, N | F = ma | kg m s−2 |
| joule, J | W = Fs | kg m2 s−2 |
| watt, W | P = E/t | kg m2 s−3 |
| pascal, Pa | p = F/A | kg m−1 s−2 |
| coulomb, C | Q = It | A s |
| volt, V | V = P/I | kg m2 s−3 A−1 |
| ohm, Ω | R = V/I | kg m2 s−3 A−2 |
You do not need to memorise the last column. Build it each time from a simple equation, as in the "Comes from" column.
How do you check an equation with units?
Both sides of a correct equation have the same units. Replace each quantity by its base units and simplify.
Show that the unit of the gravitational constant \(G\) can be written as m3 kg−1 s−2.
Rearrange \(F = \dfrac{Gm_1m_2}{r^2}\) to \(G = \dfrac{Fr^2}{m_1m_2}\).
Units: \(\dfrac{\text{N}\,\text{m}^2}{\text{kg}^2} = \dfrac{(\text{kg m s}^{-2})\,\text{m}^2}{\text{kg}^2} = \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\)
Every step is shown, which is what a "show that" question needs.
A unit check catches a wrong power or a missing quantity. It cannot catch a missing number: \(s = ut + at^2\) has the right units on both sides, but the correct equation has \(\tfrac12 at^2\).
How should units be written?
- Leave a space between the number and the unit: 9.81 m s−2.
- Negative powers (m s−1) and a solidus (m/s) are both fine. Do not mix them in one unit, and never write "mps" or "kph".
- Temperature in kelvin is written K, not °K. A temperature difference is the same in K and °C.
- Give the answer in the unit the question asks for. If it asks for kJ or cm, convert at the end.
- Label table headings and graph axes as quantity / unit, for example "time / s", so the numbers in the column are pure numbers.
How do significant figures and uncertainties fit together?
An uncertainty is itself an estimate, so it is normally quoted to one significant figure. The value is then rounded to the same decimal place as the uncertainty.
A data analysis gives \(g = 9.7836\ \text{m s}^{-2}\) with an absolute uncertainty of \(0.2184\ \text{m s}^{-2}\).
Uncertainty to 1 s.f.: 0.2. Value to the same decimal place: 9.8.
Result: \(g = (9.8 \pm 0.2)\ \text{m s}^{-2}\). Writing (9.7836 ± 0.2) m s−2 would claim digits that the uncertainty has already made meaningless.
Exam tips
Write the long calculator value on one line and the rounded answer with its unit on the next. Anyone checking your work can see the method, and you can see whether rounding was the only change.
Count the significant figures in each value you used. The least precise one decides the answer, not the one you happen to notice first.
An answer without a unit is incomplete. Put the unit on the final line every time, even when the question already names it.
When you are unsure which equation applies, check which one gives the unit you need. If the question asks for a power in watts, the answer must reduce to kg m2 s−3.
A wavelength of 589 nm put into an equation as 589 gives an answer that is wrong by a factor of 109. Convert every prefix to standard form before you substitute.
Common misconceptions
| Misconception | Why it is wrong | Correct idea |
|---|---|---|
| "More decimal places make an answer more accurate." | Extra digits from the calculator were never measured. | Match the precision of the data: the least precise value, or one more significant figure. |
| "Zeros never count." | Only leading zeros are placeholders. | Zeros between digits and trailing zeros after a decimal point are significant. |
| "If the units match, the equation is right." | Numbers such as ½ or 2π have no units. | A unit check can show an equation is wrong, never that it is right. |
| "1 cm² = 10⁻² m²." | The prefix is squared along with the unit. | 1 cm² = (10⁻² m)² = 10⁻⁴ m². |
Key rules
| Rule | Example |
|---|---|
| × and ÷: match the least precise data value | 12.4 m ÷ 3.1 s = 4.0 m s−1 |
| + and −: match the fewest decimal places | 12.4 cm + 0.356 cm = 12.8 cm |
| Round once, at the end | p = 7.39 × 103 Pa, not 7.35 × 103 Pa |
| Exact numbers do not limit precision | ½, 2π, 10 oscillations |
| Convert prefixes before substituting | 589 nm = 5.89 × 10−7 m |
| Units must match on both sides | G: m3 kg−1 s−2 |
| Uncertainty to 1 s.f., value to the same decimal place | (9.8 ± 0.2) m s−2 |
Check your understanding
1. How many significant figures are there in (a) 0.00560, (b) 3.040 × 104, (c) 700?
(a) 3: the leading zeros do not count, the trailing zero does. (b) 4: count the digits before the power of ten. (c) Ambiguous: it could be 1, 2 or 3. Write 7.00 × 102 if three are meant.
2. A trolley travels 7.25 m in 2.1 s. Calculate its average speed to an appropriate number of significant figures.
\(v = 7.25 / 2.1 = 3.452\ldots\) The time has 2 s.f., so v = 3.5 m s−1.
3. Water of mass 0.350 kg is heated from 21.9 °C to 36.4 °C. The specific heat capacity of water is 4.18 × 103 J kg−1 K−1. Calculate the energy supplied.
ΔT = 36.4 − 21.9 = 14.5 K (three significant figures survive the subtraction). \(Q = mc\Delta T = 0.350 \times 4.18\times10^3 \times 14.5 = 21\,213.5\ \text{J}\). All data have 3 s.f., so Q = 2.12 × 104 J.
4. A wire has a radius of 0.25 mm. Calculate its cross-sectional area in m2.
Convert first: r = 0.25 × 10−3 m. \(A = \pi r^2 = \pi (0.25\times10^{-3})^2 = 1.96\ldots\times10^{-7}\ \text{m}^2\). The radius has 2 s.f., so A = 2.0 × 10−7 m2.
5. Express the volt in SI base units.
V = P / I. The watt is J s−1 = kg m2 s−3. Dividing by A gives kg m2 s−3 A−1.
6. A student writes s = ut + at2. Use units to test the equation, and state what the test cannot tell you.
ut: m s−1 × s = m. at2: m s−2 × s2 = m. Both terms are in metres, so the units are consistent. The test cannot detect the missing factor ½: the correct equation is \(s = ut + \tfrac12at^2\).
7. A result is calculated as 0.04637 kg with an uncertainty of 0.00218 kg. Write it correctly.
Uncertainty to 1 s.f.: 0.002 kg. Round the value to the same decimal place: 0.046 kg. (0.046 ± 0.002) kg, or (4.6 ± 0.2) × 10−2 kg.