Chapter 1 · Units and Measurement

1.2 Significant Figures and Rounding

A measured number carries its reliability in its digits. Significant figures say how many digits we can trust, and the rules for arithmetic and rounding stop a calculator from inventing precision.

Measure a pencil with a school ruler and you might write $6.43\ \text{cm}$. Divide that by $3$ on a calculator and the display shows $2.143333333$. Do we really know the third of a pencil to a billionth of a centimetre? Of course not: the ruler could only tell us the length to about a tenth of a millimetre at best, and no amount of arithmetic can improve on the instrument.

Every measured value has limited reliability, and the way scientists signal that reliability is through the number of digits they write. This section gives the rules: how to count significant figures, how many to keep after multiplying, dividing, adding or subtracting, and how to round off fairly. From now on, every numerical answer in this course obeys them.

Significant figures

When you read a scale you write down every digit you are sure of, plus one more digit that you estimate between the markings. With a millimetre ruler, the pencil's end lies between $6.4\ \text{cm}$ and $6.5\ \text{cm}$; you judge it to be about three-tenths of the way, and write $6.43\ \text{cm}$. The $6$ and the $4$ are certain; the $3$ is uncertain but still meaningful.

📘 Definition 1.4 — Significant figures
The significant figures of a measured value are all the digits known reliably plus the first uncertain digit. The more significant figures, the more precise the measurement. The smallest division of the instrument's scale is called its least count; the estimated last digit is a fraction of the least count.
📐 Rule 1.2 — Counting significant figures
  1. All non-zero digits are significant: $3.716$ has $4$.
  2. Zeros between non-zero digits are significant: $30.05$ has $4$.
  3. Leading zeros (to the left of the first non-zero digit) are not significant — they only locate the decimal point: $0.00340$ has $3$.
  4. Trailing zeros in a number with a decimal point are significant: $4.300$ has $4$, and $0.00340$ ends in a significant zero.
  5. Trailing zeros in a whole number without a decimal point are ambiguous: "$4300\ \text{m}$" may have $2$, $3$ or $4$. Remove the doubt with scientific notation: $4.3\times10^3\ \text{m}$ ($2$), $4.300\times10^3\ \text{m}$ ($4$).
  6. In scientific notation $a\times10^b$, only the digits of $a$ count; the power of ten never adds or removes significant figures.

A consequence of rule 6: changing the unit does not change the number of significant figures. $6.43\ \text{cm}$ $=0.0643\ \text{m}$ $=64.3\ \text{mm}$ $=6.43\times10^{-2}\ \text{m}$ — always $3$, because the precision of the measurement has not changed. Counted numbers (“$20$ oscillations”) and defined numbers ($\pi$, the $2$ in $2\pi r$, $1\ \text{km}=1000\ \text{m}$) are exact: they have unlimited significant figures and never limit an answer.

✏️ Example 1 — Counting significant figures
How many significant figures are there in (a) $0.007020$, (b) $4.300\times10^3$, (c) $30.050$, (d) $6.02\times10^{23}$, (e) $0.40$, (f) $4300$?
(a) Leading zeros $0.00$ do not count; $7$, the sandwiched $0$, $2$ and the trailing $0$ (after a decimal point) do: $4$.
(b) Count the digits of $4.300$: $4$.
(c) Every digit counts (sandwiched zeros and a trailing zero after the point): $5$.
(d) $6.02$: $3$. The $10^{23}$ adds nothing.
(e) $0.40$: $2$ — writing the final zero claims the hundredths digit is known to be $0$.
(f) Ambiguous: between $2$ and $4$. A careful writer would use $4.3\times10^3$, $4.30\times10^3$ or $4.300\times10^3$.
Interactive 1 — Reading a Scale: Certain Digits and One Estimated Digit

A rod lies along a ruler; the magnified window shows its right end. Choose a ruler with centimetre or millimetre markings. The reading keeps every digit fixed by the markings and adds one estimated digit — a tenth of the least count. Press Play to let the rod's length drift and watch the reading and its number of significant figures.

certain digits—
reading (cm)—
significant figures—
same reading in metres—

Significant figures in arithmetic

A result cannot be more reliable than the least reliable data that produced it. How that idea is applied depends on the operation.

📐 Rule 1.3 — Arithmetic with significant figures

Multiplication and division: the result keeps as many significant figures as the factor with the fewest significant figures.

Addition and subtraction: the result keeps as many decimal places as the term with the fewest decimal places (all terms in the same unit).

Exact numbers are ignored when applying both rules. In a calculation with several steps, keep one or two extra digits in the intermediate results and round only the final answer.

Why two different rules? In a product, a relative (fractional) uncertainty in one factor becomes the same relative uncertainty in the answer, so what matters is how many figures each factor has. In a sum, an uncertainty in the hundredths place of one term makes the hundredths place of the total uncertain, however many figures the other terms have — the position of the last reliable digit matters.

✏️ Example 2 — Area of a floor
A rectangular floor measures $3.716\ \text{m}$ by $2.08\ \text{m}$. Find its area.
Calculator $3.716\times2.08=7.72928$.
Count $3.716$ has $4$ significant figures, $2.08$ has $3$. Keep $3$.
Answer $A=7.73\ \text{m}^2$.
Why The width is only known to about $\pm0.01\ \text{m}$, i.e. about $0.5\%$; $0.5\%$ of $7.73$ is about $0.04$, so even the hundredths digit of the area is uncertain. Reporting $7.72928$ would claim a precision of a thousandth of a percent.
✏️ Example 3 — Adding masses
A beaker of mass $352.71\ \text{g}$ (measured on a laboratory balance) contains $18.6\ \text{g}$ of salt (weighed on a kitchen scale) and a $0.247\ \text{g}$ stirrer (weighed on an analytical balance). What is the total mass?
Add $352.71+18.6+0.247=371.557\ \text{g}$.
Decimal places $2$, $1$ and $3$ decimal places. The kitchen scale fixes only the tenths of a gram, so the sum is known only to the tenths: keep $1$ decimal place.
Answer $371.6\ \text{g}$ — four significant figures, even though one term had only three. In addition the count of significant figures can go up or down; only decimal places matter.
✏️ Example 4 — Density and an exact number
(a) A metal sample of mass $7.86\ \text{g}$ has volume $2.9\ \text{cm}^3$. Find its density. (b) A cube has side $4.12\ \text{cm}$. Find its total surface area and volume.
(a) $\rho=\dfrac{7.86}{2.9}=2.7103\ldots\ \text{g/cm}^3$. The volume has only $2$ significant figures: $\rho=2.7\ \text{g/cm}^3$.
(b) Area $A=6a^2=6\times(4.12)^2=101.8464\ \text{cm}^2$. The $6$ (six faces) is exact, so it does not limit the answer; $4.12$ has $3$ figures: $A=102\ \text{cm}^2$.
(b) Volume $V=a^3=69.934528\ \text{cm}^3\to69.9\ \text{cm}^3$.
Guard digits If a later step needs $V$ — say a density with mass $152.6\ \text{g}$ — use $69.93$ (one extra digit), not $69.9$: $\tfrac{152.6}{69.93}=2.182\to2.18\ \text{g/cm}^3$.

Rounding off

After deciding how many figures to keep, we must drop the rest. The rules are designed so that, on average, rounding pushes values up as often as down.

📐 Rule 1.4 — Rounding off
Look at the part to be dropped.
  1. If it is more than half a unit of the last kept digit (first dropped digit $\gt5$, or $5$ followed by any non-zero digit), increase the last kept digit by $1$: $2.746\to2.75$, $1.0051\to1.01$.
  2. If it is less than half (first dropped digit $\lt5$), leave the last kept digit unchanged: $2.743\to2.74$.
  3. If it is exactly half (a single $5$, or $5$ followed only by zeros), make the last kept digit even: leave it if it is already even, increase it by $1$ if it is odd. $7.385\to7.38$ ($8$ is even); $7.375\to7.38$ ($7$ is odd).
✏️ Example 5 — Rounding to three significant figures
Round to $3$ significant figures: (a) $15.348$, (b) $0.072849$, (c) $15.35$, (d) $15.25$, (e) $299\,792\,458$.
(a) Drop $48$: less than half: $15.3$.
(b) The first three significant digits are $7,2,8$; the dropped part $49$ is less than half: $0.0728$.
(c) Exactly half; last kept digit $3$ is odd → $15.4$.
(d) Exactly half; last kept digit $2$ is even → $15.2$.
(e) $2.99792458\times10^8$: dropped part $792\ldots$ is more than half, so $2.99\to3.00$: $3.00\times10^8$ — the zeros must be written to show three figures.
⚠️ Common mistakes

1. Rounding at every step. Rounding intermediate results makes errors pile up. Keep a guard digit and round once at the end.

2. Copying the calculator. "$\rho=2.710344828\ \text{g/cm}^3$" is wrong, not "more accurate".

3. Dropping significant zeros. If the answer is $4.00\ \text{m/s}$ to three figures, writing "$4\ \text{m/s}$" throws information away.

4. Using the wrong rule. Decimal places for $+$ and $-$; significant figures for $\times$ and $\div$. In a mixed expression, apply the rules step by step in the order of the operations.

Interactive 2 — Which Digits Survive?

Each data digit is coloured: significant (purple), the uncertain last digit (amber box), and leading zeros that only place the point (grey). The calculator answer is then cut back by Rule 1.3 — significant figures for $\times$ and $\div$, decimal places for $+$ and $-$ (uncertain columns shaded) — and rounded by Rule 1.4. Press Play to step through examples, or deal new random numbers.

calculator says—
rule applied—
correct answer—
🚀 Beyond the board syllabus (JEE Main · NEET · JEE Advanced) — errors in measurement

Board papers no longer test this, but the JEE Main, NEET and JEE Advanced syllabuses still list errors in measurement (JEE Advanced as error analysis in experiments). Read it once now; it explains why the significant-figure rules work.

Absolute, relative and percentage error. If repeated measurements give $a_1,a_2,\ldots,a_n$, the best value is the mean $\bar a=\tfrac1n\sum a_i$. The absolute errors are $\Delta a_i=|a_i-\bar a|$ and the mean absolute error is $\Delta a=\tfrac1n\sum\Delta a_i$; we report $a=\bar a\pm\Delta a$. The relative error is $\dfrac{\Delta a}{\bar a}$ and the percentage error is $\dfrac{\Delta a}{\bar a}\times100\%$.

Combining errors (maximum-error rules):

  • Sum or difference, $Z=A\pm B$: absolute errors add, $\Delta Z=\Delta A+\Delta B$.
  • Product or quotient, $Z=AB$ or $Z=\dfrac AB$: relative errors add, $\dfrac{\Delta Z}{Z}=\dfrac{\Delta A}{A}+\dfrac{\Delta B}{B}$.
  • Powers, $Z=\dfrac{A^pB^q}{C^r}$: $\dfrac{\Delta Z}{Z}=p\dfrac{\Delta A}{A}+q\dfrac{\Delta B}{B}+r\dfrac{\Delta C}{C}$ — the quantity with the highest power dominates.

Worked example — $g$ from a pendulum. Five timings of one period give $1.98,\ 2.03,\ 2.01,\ 1.96,\ 2.02\ \text{s}$. Mean $\bar T=\tfrac{10.00}{5}=2.000\ \text{s}$; absolute errors $0.02,\,0.03,\,0.01,\,0.04,\,0.02$, mean $0.024\approx0.02\ \text{s}$; so $T=2.00\pm0.02\ \text{s}$, a $1.0\%$ error. The length is $l=100.0\pm0.1\ \text{cm}$, a $0.1\%$ error. From $g=\dfrac{4\pi^2l}{T^2}$: $g=\dfrac{4\pi^2(1.000)}{(2.00)^2}=9.87\ \text{m/s}^2$ and $\dfrac{\Delta g}{g}=0.1\%+2(1.0\%)=2.1\%$, so $\Delta g\approx0.2\ \text{m/s}^2$ and $g=9.9\pm0.2\ \text{m/s}^2$. The $2.1\%$ uncertainty is why only two significant figures survive — the same conclusion Rule 1.3 reaches by counting digits.

Exercises

I
Counting significant figures.
  • 1 $0.0503$
  • 2 $2.0500$
  • 3 $6.320\times10^4$
  • 4 $1.002$
  • 5 $0.00080$
  • 6 $5000\ \text{kg}$ — and how would you write it to show exactly $2$ significant figures?
  • 7 $5.000\times10^3\ \text{kg}$
  • 8 Express $2.70\ \text{m}$ in centimetres and in millimetres without changing the number of significant figures.
1 $3$ (the leading zeros do not count; the sandwiched $0$ does).
2 $5$.
3 $4$.
4 $4$.
5 $2$ (the $8$ and the final $0$).
6 Ambiguous — from $1$ to $4$. Two figures: $5.0\times10^3\ \text{kg}$.
7 $4$.
8 $2.70\times10^2\ \text{cm}$ (writing "$270\ \text{cm}$" would be ambiguous) and $2.70\times10^3\ \text{mm}$.
II
Rounding. Round each number to $3$ significant figures.
  • 1 $2.4567$
  • 2 $0.072849$
  • 3 $8.135$
  • 4 $8.125$
  • 5 $1.0051$
  • 6 $6.0249\times10^{-5}$
1 Dropped $67$ > half: $2.46$.
2 $0.0728$.
3 Exactly half, $3$ odd → $8.14$.
4 Exactly half, $2$ even → $8.12$.
5 Dropped part $51$ is more than half (a $5$ followed by a non-zero digit): $1.01$.
6 Dropped $49$ < half: $6.02\times10^{-5}$.
III
Arithmetic. Give each answer to the correct number of significant figures, with its unit.
  • 1 $2.31\ \text{m}\times0.45\ \text{m}$
  • 2 $12.6\ \text{m}\div3.15\ \text{s}$
  • 3 $23.452\ \text{g}+1.2\ \text{g}+0.03\ \text{g}$
  • 4 $9.04\ \text{cm}-8.9\ \text{cm}$
  • 5 $(4.5\times10^3\ \text{N})\times(2.250\times10^{-2}\ \text{m})$
  • 6 The area of a circle of radius $2.40\ \text{cm}$.
  • 7 A sprinter covers $100.0\ \text{m}$ in $9.58\ \text{s}$. Find the mean speed.
  • 8 The perimeter of a rectangle $12.35\ \text{cm}$ by $4.2\ \text{cm}$.
1 $1.0395\to1.0\ \text{m}^2$ ($2$ figures, from $0.45$).
2 $4\to4.00\ \text{m/s}$ ($3$ figures — the zeros must be written).
3 $24.682\to24.7\ \text{g}$ ($1$ decimal place).
4 $0.14\to0.1\ \text{cm}$ ($1$ decimal place). Subtracting nearly equal numbers destroys precision: only $1$ figure is left.
5 $101.25\to1.0\times10^2\ \text{N m}$ ($2$ figures).
6 $\pi(2.40)^2=18.096\to18.1\ \text{cm}^2$ ($\pi$ is exact; use enough of its digits).
7 $\tfrac{100.0}{9.58}=10.438\to10.4\ \text{m/s}$.
8 $12.35+4.2=16.55$ (keep a guard digit), $\times2$ (exact) $=33.10\to33.1\ \text{cm}$ ($1$ decimal place, from $4.2$).
IV
Applications.
  • 1 Engineering. A steel sheet is $2.465\ \text{m}$ long, $1.25\ \text{m}$ wide and $0.0040\ \text{m}$ thick; steel has density $7.85\times10^3\ \text{kg/m}^3$. Find the mass of the sheet.
  • 2 Lab. The time for $20$ oscillations of a pendulum is $36.4\ \text{s}$. Find the period, and explain why the answer has three significant figures.
  • 3 A student finds a density as $\tfrac{13.5\ \text{g}}{4.97\ \text{cm}^3}$ and writes $2.716297787\ \text{g/cm}^3$. Correct the answer and explain.
  • 4 Biology. A cell is measured as $0.0125\ \text{mm}$ across. Write this in micrometres and in metres, keeping the number of significant figures.
  • 5 Sport. A runner's three lap times are $52.3\ \text{s}$, $51.86\ \text{s}$ (from a different stopwatch) and $53.1\ \text{s}$. Find the total time to the correct precision.
1 $V=2.465\times1.25\times0.0040=0.012325\ \text{m}^3$; $m=\rho V=96.75\ \text{kg}\to97\ \text{kg}$ (the thickness has only $2$ figures).
2 $T=\tfrac{36.4}{20}=1.82\ \text{s}$. The $20$ is a count (exact), so the $3$ figures of $36.4$ decide.
3 $2.72\ \text{g/cm}^3$: both data have $3$ figures, so the answer has $3$. The extra digits are not information.
4 $0.0125\ \text{mm}=12.5\ \text{µm}=1.25\times10^{-5}\ \text{m}$ — three significant figures each time.
5 $52.3+51.86+53.1=157.26\ \text{s}\to157.3\ \text{s}$ ($1$ decimal place, set by the tenth-of-a-second readings).
🎯 Exam corner BoardJEE MainJEE AdvancedNEET

E1. Board · 3 marks State the rule for the number of significant figures in the result of (i) a multiplication or division, (ii) an addition or subtraction. Use them to evaluate $5.24\times3.1$ and $24.36+0.0812+256.1$.

Rules (i) As many significant figures as the factor with the fewest. (ii) As many decimal places as the term with the fewest. [1]
Product $5.24\times3.1=16.244$; $3.1$ has $2$ figures → $16$. [1]
Sum $24.36+0.0812+256.1=280.5412$; $256.1$ has $1$ decimal place → $280.5$. [1]
Exam tip Write the unrounded value first and then the rounded one — it shows the examiner you applied the rule rather than guessed.

E2. NEET The number of significant figures in $0.06900$ is (a) $5$ (b) $4$ (c) $2$ (d) $3$

Count Leading $0.0$ do not count; $6$, $9$ and both trailing zeros (after the decimal point) do.
Answer (b) $4$.
Exam tip Option (a) counts the leading zero after the point; option (c) forgets the trailing zeros. Write the number as $6.900\times10^{-2}$ and count the digits of the mantissa.

E3. JEE Main The side of a cube is measured as $1.2\times10^{-2}\ \text{m}$. Its volume, written to the correct number of significant figures, is (a) $1.728\times10^{-6}\ \text{m}^3$ (b) $1.73\times10^{-6}\ \text{m}^3$ (c) $1.7\times10^{-6}\ \text{m}^3$ (d) $1.70\times10^{-6}\ \text{m}^3$

Compute $(1.2\times10^{-2})^3=1.728\times10^{-6}\ \text{m}^3$.
Rule Cubing is repeated multiplication, and $1.2$ has $2$ figures.
Answer (c) $1.7\times10^{-6}\ \text{m}^3$.
Exam tip (d) has the right value but claims $3$ figures; the trailing zero matters.

E4. JEE Main NEET In an experiment $g$ is found from $g=\dfrac{4\pi^2l}{T^2}$. The length $l$ has a $1\%$ error and the period $T$ a $2\%$ error. The maximum percentage error in $g$ is (a) $3\%$ (b) $4\%$ (c) $5\%$ (d) $1\%$

Note Errors in measurement are an exam extension, beyond the board syllabus (see the 🚀 box).
Rule For $g\propto l^1T^{-2}$: $\dfrac{\Delta g}{g}=\dfrac{\Delta l}{l}+2\dfrac{\Delta T}{T}$.
Answer $1\%+2(2\%)=5\%$: (c).
Exam tip A negative power still adds its error; (a) forgets the factor $2$, and "$1\%$" subtracts instead of adding.

E5. JEE Advanced (One or more correct) A quantity is calculated from $P=\dfrac{a^3b^2}{\sqrt c\,d}$. The percentage errors in $a$, $b$, $c$, $d$ are $1\%$, $3\%$, $4\%$ and $2\%$, and the calculated value is $P=3.763$. Then (A) the maximum percentage error in $P$ is $13\%$ (B) $P$ should be reported as $3.763$ (C) $P$ should be reported as $3.8\pm0.5$ (D) the largest single contribution to the error comes from $b$

Note This uses the error-combination rules of the exam extension.
Contributions $a$: $3\times1\%=3\%$; $b$: $2\times3\%=6\%$; $c$: $\tfrac12\times4\%=2\%$; $d$: $1\times2\%=2\%$. Total $13\%$. (A) ✓, (D) ✓.
Reporting $13\%$ of $3.763$ is $0.49\approx0.5$, so already the first decimal place is uncertain: $P=3.8\pm0.5$. (C) ✓; (B) claims far too much ✗.
Answer (A), (C), (D).
Exam tip The error decides the significant figures: round the error to one figure, then round the value to the same decimal place.
🔬 Research–Teaching NexusResearch-tutored · optional, about 30 minutes

Research question. When is a sports timing result too precise to be meaningful? The case of the 1972 Olympic 400 m individual medley.

Why it matters. At the 1972 Munich Olympics, the gold and silver medals in the men's 400 m individual medley were separated by two thousandths of a second on the electronic clock. Afterwards swimming's governing body decided to time races only to hundredths of a second, because the pools themselves are not built that precisely. It is a real-world decision about significant figures.

What to do.

  1. The winning time was about $4\ \text{min}\ 31.98\ \text{s}$ for $400\ \text{m}$. Find the mean speed to three significant figures.
  2. How far does a swimmer travel in $0.002\ \text{s}$ at that speed? Give your answer in millimetres.
  3. Competition rules allow each lane of a $50\ \text{m}$ pool to be up to $3\ \text{cm}$ longer than nominal (and never shorter). A $400\ \text{m}$ race is $8$ lengths. What is the largest possible difference in distance between two lanes, and what time does it correspond to?
  4. Decide how many significant figures a $400\ \text{m}$ time can honestly be given to, and compare with the hundredths rule.

Tools: Calculator; the current pool-facility rules of the sport's international federation.

Hand in: A short argument (about half a page) with your numbers, ending with a recommendation on how times should be reported.

Going further. Investigate photo-finish timing in athletics (times recorded to $0.001\ \text{s}$ but published to $0.01\ \text{s}$) or the precision of ski-racing courses. Is the reporting rule the same?

Instructor note: $v=\tfrac{400}{271.98}=1.47\ \text{m/s}$; in $0.002\ \text{s}$ the swimmer moves $\approx2.9\ \text{mm}$; lane difference up to $8\times3\ \text{cm}=24\ \text{cm}$, i.e. about $0.16\ \text{s}$ at $1.47\ \text{m/s}$. A 2 ms margin is therefore far below the uncertainty of the course itself; even hundredths are generous. Good answers connect "precision of the clock" with "accuracy of the whole measurement".

🧪 Self-check quiz

0/5 correct

Q1. How many significant figures does $0.00340$ have?

Q2. $3.0\times2.115=$

Q3. $12.11\ \text{m}+18.0\ \text{m}+1.013\ \text{m}=$

Q4. Rounded to three significant figures, $6.225$ becomes

Q5. A length of $4.700\ \text{m}$ is rewritten as $470.0\ \text{cm}$. The number of significant figures

← 1.1 Physical Quantities and the International System of Units1.3 Dimensions and Dimensional Analysis →