Significant figures and errors in a result

How sure are you of a number?

You measure a pencil with a ruler and say "$14.3$ cm". A friend says "$14.30$ cm". Are these the same? No. The first says that you are sure of the tenths of a centimetre. The second claims that you know the hundredths too. A number in physics carries a promise about how well it is known.

Significant figures

The significant figures of a number are the digits that are known for sure, plus one digit that is uncertain. Here are the rules.

• All non-zero digits count: $4.327$ has four.
• Zeros between non-zero digits count: $2.006$ has four.
• Zeros at the start do not count. They only place the decimal point: $0.0450$ has three.
• Zeros at the end count when there is a decimal point: $0.0450$ has three, and $3.00$ has three.
• Zeros at the end of a whole number are doubtful: does $2300$ have two, three or four? Write it in the scientific form to show: $2.3 \times 10^3$ (two), $2.30 \times 10^3$ (three).
• Exact numbers, like $20$ oscillations or $2$ in $2\pi r$, have no uncertainty.
• Changing units does not change the count: $14.3$ cm is $0.143$ m, and both have three.

Adding and subtracting. The answer has as many decimal places as the number with the fewest decimal places. For example, $12.11 + 18.0 + 1.013$ $= 31.123$, which we write as $31.1$.

Multiplying and dividing. The answer has as many significant figures as the number with the fewest. For example, $4.327 \times 2.1 = 9.0867$, which we write as $9.1$.

Rounding. If the digit to drop is less than 5, leave the last digit. If it is more than 5, add 1. If it is exactly 5, NCERT makes the last digit even: $2.45$ becomes $2.4$ and $2.35$ becomes $2.4$. Many examinations, and most calculators, round a 5 up. Say which one you use. Round only at the end of the calculation.

The error of one reading

A single reading can be wrong by about the least count of the instrument. A ruler with 1 mm marks gives an error of about $\pm 1$ mm. The vernier calipers give $\pm 0.1$ mm, and a screw gauge gives $\pm 0.01$ mm. A stopwatch reads to $0.1$ s or $0.01$ s. A-Level and IB books often take half the smallest division of an analogue scale.

Absolute, relative and percentage error

Take several readings $a_1, a_2, \ldots, a_n$ of the same quantity.

1. The mean is the best value: $a_{mean} = \dfrac{a_1 + a_2 + \cdots + a_n}{n}$.
2. The absolute error of each reading is $\Delta a_i = |a_{mean} - a_i|$. It has the unit of the quantity.
3. The mean absolute error is $\Delta a_{mean} = \dfrac{\Delta a_1 + \cdots + \Delta a_n}{n}$. This is the error that we write with the result: $a = a_{mean} \pm \Delta a_{mean}$.
4. The relative error is $\dfrac{\Delta a_{mean}}{a_{mean}}$. It has no unit.
5. The percentage error is $\dfrac{\Delta a_{mean}}{a_{mean}} \times 100\%$.

If the mean absolute error is smaller than the least count, use the least count instead. The error of a result is never smaller than what the instrument can show.

A worked example. A student measures the length of a block five times with vernier calipers of least count $0.01$ cm: $5.23$, $5.25$, $5.22$, $5.24$ and $5.26$ cm. Find the mean, the mean absolute error and the percentage error.

1. The mean. The five readings add up to $26.20$ cm. So the mean is $\dfrac{26.20\ \text{cm}}{5} = 5.24$ cm.
2. The absolute errors. $0.01$, $0.01$, $0.02$, $0.00$ and $0.02$ cm.
3. The mean absolute error. The five errors add up to $0.06$ cm. So the mean is $\dfrac{0.06\ \text{cm}}{5} = 0.012$ cm. This is just above the least count, and we write it as $0.01$ cm.
4. The result. $L = 5.24 \pm 0.01$ cm. The relative error is $\dfrac{0.012}{5.24} = 0.0023$, which is $0.23\%$.

Combining errors

The rules follow if each reading is at its worst. They give the maximum error.

• A sum or a difference, $Z = A \pm B$: the absolute errors add, $\Delta Z = \Delta A + \Delta B$. Even for a difference, the errors add. Subtracting two nearly equal numbers loses accuracy.
• A product or a quotient, $Z = A \times B$ or $Z = A/B$: the relative errors add, $\dfrac{\Delta Z}{Z} = \dfrac{\Delta A}{A} + \dfrac{\Delta B}{B}$.
• A power, $Z = A^p B^q$: $\dfrac{\Delta Z}{Z} = p\,\dfrac{\Delta A}{A} + q\,\dfrac{\Delta B}{B}$. The power multiplies the relative error, and the sign of the power does not matter.

Set the powers to $p = 1$ and $q = 1$. Then raise $p$ to $4$. How does the share of A in the error of Z change?

A–Level, IB and statistics. Their books call the absolute error the uncertainty, and write the percentage uncertainty in the same way. They add the percentage uncertainties for a product, as we do. For independent random errors, university courses add them in quadrature, $\sqrt{(\Delta A)^2 + (\Delta B)^2}$, which gives a smaller value. JEE and the board examinations use the simple sum.

A worked example. The density of a metal cylinder is $\rho = \dfrac{4m}{\pi d^2 L}$. The mass is $71.4 \pm 0.1$ g, the diameter is $1.20 \pm 0.01$ cm and the length is $7.52 \pm 0.01$ cm. Find the density and its error.

1. The density. $\rho$ $= \dfrac{4 \times 71.4\ \text{g}}{\pi \times (1.20\ \text{cm})^2 \times 7.52\ \text{cm}}$ $= 8.40\ \text{g/cm}^3$.
2. The relative errors. $\dfrac{\Delta m}{m} = \dfrac{0.1}{71.4} = 0.0014$, and $\dfrac{\Delta L}{L} = \dfrac{0.01}{7.52} = 0.0013$. The diameter is squared: $2\,\dfrac{\Delta d}{d} = 2 \times \dfrac{0.01}{1.20} = 0.0167$.
3. The sum. $\dfrac{\Delta \rho}{\rho}$ $= 0.0014 + 0.0167 + 0.0013$ $= 0.0194$, which is $1.9\%$.
4. The absolute error. $\Delta \rho = 0.0194 \times 8.40\ \text{g/cm}^3$ $= 0.16\ \text{g/cm}^3$. We write $\rho = 8.4 \pm 0.2\ \text{g/cm}^3$.

The diameter causes most of the error, because it is squared and because it is the smallest length. To improve the experiment, measure the diameter with a better instrument. A better balance would help very little.

Practise with new numbers

Try it in Python

The program does the density calculation of the worked example from the tables of readings. It uses the mean deviation, but never less than the least count.

Common mistakes

• Subtracting the errors of a difference. The absolute errors add, for a sum and for a difference.
• Forgetting the power. In $d^2$ the relative error of $d$ counts twice.
• Writing more digits than the instrument gives. $5.2400000$ cm from a ruler is not an honest result.
• Rounding in the middle. Keep an extra digit until the last step.
• Writing the error with many digits. An error is written with one significant figure, rarely two. The result then has the same decimal places.
• Taking an error smaller than the least count. If the readings agree exactly, the error is still about the least count.

Remember these

• Significant figures: count from the first non-zero digit. In a sum keep the fewest decimal places. In a product keep the fewest significant figures.
• Absolute error $\Delta a$, relative error $\Delta a / a$, percentage error $\dfrac{\Delta a}{a} \times 100\%$.
• Sum or difference: add the absolute errors. Product or quotient: add the relative errors. A power $p$ multiplies the relative error by $p$.
• The error of a result is not smaller than the least count of the instrument.
• Write a result as $a_{mean} \pm \Delta a$ with its unit.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. How many significant figures do $0.00600$, $6.0 \times 10^4$ and $60.02$ have?

2. Two lengths are $25.4 \pm 0.1$ cm and $12.6 \pm 0.1$ cm. Find their difference with its error.

3. The radius of a sphere is $2.50 \pm 0.02$ cm. What is the percentage error in its volume?

4. A stopwatch with a least count of $0.1$ s times 20 oscillations as $36.0$ s. What is the error in the time period $T$?

References

• NCERT, Physics Class XI, Chapter 1 (Units and Measurements), the sections on significant figures and on errors.
• OpenStax University Physics Volume 1, Chapter 1, the section on significant figures.