Significant figures in physical results

Significant figures are the digits in a recorded value that carry information about its accuracy; counting begins at the first non-zero digit on the left. Leading zeros are not significant ($0.0045$ has two significant figures), internal zeros are significant ($1.02$ has three), and trailing zeros are significant only when a decimal part is present ($2.30$ has three, but $2300$ with no further notation has two). In standard form $a\times 10^{n}$ with $1\le a<10$, the exponent $n$ does not affect precision: all digits of the factor $a$ count. In multiplication and division the result is rounded to the smallest number of significant figures among the inputs, while in addition and subtraction it is rounded to the smallest number of decimal places. Intermediate steps are carried with one or two extra digits and only the final answer is rounded, otherwise rounding error accumulates. Significant-figure rules give a quick estimate of precision; a stricter method is to state the uncertainty directly, for example $L=(12.35\pm 0.05)$ cm.

Determine the number of significant figures in: (a) $0.004507$ m; (b) $2.30\times10^{4}$ kg; (c) $1500$ m (no decimal part, trailing zeros not specified).

Find the area of a rectangle with sides $a=12.35$ cm and $b=4.2$ cm and write the answer with the correct number of significant figures. Then add $12.35$ cm and $0.2$ cm and round the sum by the addition rule.

The density of a metal is determined as follows: mass $m=24.31$ g (four significant figures), volume by displacement with initial level $V_{1}=50.0$ mL and final level $V_{2}=58.5$ mL. Compute $\rho=m/(V_{2}-V_{1})$, round the result correctly, and explain why writing extra digits is misleading.

Measurement and rounding (safe, at home). Take a book and a ruler with a scale division of $1$ mm. Measure the length and width of the book three times, shifting the ruler slightly each time, and record the results to $0.1$ cm. For each pair of measurements compute the area, round it by the significant-figure rules, and arrange everything in a table: measurement, number of significant figures, area before rounding, area after rounding. Then answer in writing: how many significant figures does each measurement give, and which one limits the precision of the area? If no ruler or object is available, carry out the same calculation in an interactive measurement simulation (virtual ruler or vernier caliper): it is safe and needs no instruments. Do not use open flames, sharp tools, or energised electrical circuits.

To show which digits in a recorded value count as significant and to recall which rounding rule applies to each operation.

Sketch: the value $0.004507$ with labels and arrows — leading zeros marked as not significant, digits $4$, $5$, $0$, $7$ marked as significant. Next to it a flow chart with two branches: "multiplication/division $\rightarrow$ smallest number of significant figures" and "addition/subtraction $\rightarrow$ smallest number of decimal places", with the rounding examples $51.87\rightarrow 52$ and $12.55\rightarrow 12.6$ below.

At a school competition the distance was measured with a tape: $s = 100.00\ \text{m}$. The electronic timer showed $t = 12.34\ \text{s}$. The calculator gives $v = 100.00 / 12.34 = 8.103726\ldots\ \text{m/s}$. The official result is written as $v = 8.10\ \text{m/s}$, although the calculator screen shows many more digits.

This is the division rule: in multiplication and division the result keeps as many significant figures as the least precise input. The time $12.34\ \text{s}$ has four, the distance $100.00\ \text{m}$ has five, so the answer keeps four: $8.104\ \text{m/s}$. Writing $8.1037\ \text{m/s}$ would credit the timer with a precision of $0.0001\ \text{s}$ that it does not have.

A digital kitchen scale has a scale division of $0.001\ \text{kg}$. Flour for dough is weighed out and the display reads $0.500\ \text{kg}$. The same value written as $0.5\ \text{kg}$ implies a different scale with a division of $0.1\ \text{kg}$, while $0.5000\ \text{kg}$ would require a scale with a division of $0.0001\ \text{kg}$.

Zeros after the decimal point at the end of a number are significant: $0.500\ \text{kg}$ states that the mass lies between $0.4995\ \text{kg}$ and $0.5005\ \text{kg}$, that is, it carries three significant figures. In this way the record of a result ties the number of digits to the scale division of the instrument and to the real uncertainty of the weighing.

In the morning the outdoor thermometer read $t_1 = -3.4\ \text{°C}$, in the afternoon $t_2 = +5.2\ \text{°C}$. The warming during the day is $\Delta t = t_2 - t_1 = 5.2 - (-3.4) = 8.6\ \text{°C}$. Writing $8.60\ \text{°C}$ would be wrong, even though both readings carry two significant figures each.

This is the addition and subtraction rule: the result keeps not the number of significant figures but the number of decimal places — the smallest among the terms. Both readings have one digit after the decimal point, so the difference has one as well: $8.6\ \text{°C}$. The added zero in $8.60$ would claim a thermometer precision of $0.01\ \text{°C}$.