Measuring instruments: range and precision

Key idea: A useful measurement needs a suitable range and scale, a zero check and a careful reading. Repeats reduce random variation; timing many cycles improves a period estimate.

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A measuring tape is ideal for a bench but useless for the thickness of a wire, and a stopwatch that shows hundredths of a second cannot beat your reaction time. A good measurement starts with an instrument whose range and precision suit the job. It then depends on reading the scale correctly, checking the zero, timing many oscillations, repeating readings and measuring many items to find one.

Range and precision

Two properties decide whether an instrument suits a measurement.

  • The range runs from the smallest to the largest value the instrument can measure. A 30 cm rule measures lengths up to 30 cm; a 0–1 A ammeter measures currents up to 1 A.
  • The resolution describes the smallest change the instrument can distinguish. For these school instruments, compare the smallest scale division or display step: a metre rule marked in millimetres has 1 mm divisions, while the stated digital calipers change in 0.01 mm steps. Fine resolution helps you distinguish small changes. Precision is how closely repeated measurements agree; accuracy is how close a measurement is to the true value. See the measurement vocabulary’s distinction.

Choose in this order:

  1. Range: the range must cover the value you expect.
  2. Scale or display step: one division must be small compared with the value, and with any difference you need to detect.
  3. Practicality: the instrument must fit the object and be quick to use well.

A tape with 1 mm divisions can measure a 1.20 m bench as 1200 mm. A 1 mm division is tiny compared with this length. The same 1 mm is bigger than a wire 0.5 mm thick, so a rule cannot tell a 0.4 mm wire from a 0.6 mm one.

Instruments you should know

Typical ranges, resolutions and uses
InstrumentQuantityTypical rangeTypical resolutionSuits
Measuring tapeLengthTens of metres1 mmA room or corridor
Metre ruleLength0–100 cm1 mmA few centimetres to 1 m
Digital calipersLength0–150 mm0.01 mmThe diameter of a rod or marble; the inside diameter of a tube
Digital micrometerLength0–25 mm0.001 mmThe diameter of a wire; the thickness of a thin sheet
Digital stopwatchTimeHours0.01 s shownTiming by hand
Measuring cylinderVolume10 cm³ to 1000 cm³0.2 cm³ to 10 cm³A liquid; a solid, by displacement
Electronic balanceMassA few hundred grams0.1 g or 0.01 gSmall masses
Spring balance (newton meter)Weight0–10 N0.2 NA force or weight
Laboratory thermometerTemperature−10 °C to 110 °C1 °CLiquids in the laboratory
AmmeterCurrent0–1 A0.02 ACurrent through a component
VoltmeterPotential difference0–3 V or 0–5 V0.1 Vp.d. across a component

Choose a quantity to compare its instruments.

Length4 instruments

Measuring tape

Typical range
Tens of metres
Typical resolution
1 mm

Suits: A room or corridor

Metre rule

Typical range
0–100 cm
Typical resolution
1 mm

Suits: A few centimetres to 1 m

Digital calipers

Typical range
0–150 mm
Typical resolution
0.01 mm

Suits: The diameter of a rod or marble; the inside diameter of a tube

Digital micrometer

Typical range
0–25 mm
Typical resolution
0.001 mm

Suits: The diameter of a wire; the thickness of a thin sheet

Time1 instrument

Digital stopwatch

Typical range
Hours
Typical resolution
0.01 s shown

Suits: Timing by hand

Volume1 instrument

Measuring cylinder

Typical range
10 cm³ to 1000 cm³
Typical resolution
0.2 cm³ to 10 cm³

Suits: A liquid; a solid, by displacement

Mass1 instrument

Electronic balance

Typical range
A few hundred grams
Typical resolution
0.1 g or 0.01 g

Suits: Small masses

Weight1 instrument

Spring balance (newton meter)

Typical range
0–10 N
Typical resolution
0.2 N

Suits: A force or weight

Temperature1 instrument

Laboratory thermometer

Typical range
−10 °C to 110 °C
Typical resolution
1 °C

Suits: Liquids in the laboratory

Current1 instrument

Ammeter

Typical range
0–1 A
Typical resolution
0.02 A

Suits: Current through a component

Potential difference1 instrument

Voltmeter

Typical range
0–3 V or 0–5 V
Typical resolution
0.1 V

Suits: p.d. across a component

The ranges and resolutions are typical values. Read them from the instrument you actually use.

Worked example 1

Choose and justify an instrument

Problem

Choose an instrument for each measurement and explain why: the length of a laboratory bench, about 1.5 m; the diameter of a marble, about 16 mm; the diameter of a copper wire, about 0.3 mm; 8 cm³ of a liquid.

Show full solution
  1. Bench, about 1.5 m

    Method

    Choose a measuring tape.

    Reason

    Its range of tens of metres covers 1.5 m in one go, and 1 mm is tiny compared with 1500 mm.

    Working

    Answer

    A 30 cm rule would have to be laid down five times end to end, adding a reading error each time.

  2. Marble, about 16 mm

    Method

    Choose digital calipers.

    Reason

    16 mm is inside their range, and 0.01 mm is small compared with 16 mm.

    Working

    Answer

    A rule’s 1 mm division is one-sixteenth of the diameter, and a rule is hard to line up with a curved surface.

  3. Copper wire, about 0.3 mm

    Method

    Choose a digital micrometer.

    Reason

    Its 0.001 mm display step gives a reading such as 0.312 mm.

    Working

    Answer

    Calipers would give only 0.31 mm, and a rule could not measure it at all.

  4. 8 cm³ of a liquid

    Method

    Choose a 10 cm³ measuring cylinder with 0.2 cm³ divisions.

    Reason

    It covers 8 cm³, and 0.2 cm³ is small compared with 8 cm³.

    Working

    Answer

    A 100 cm³ cylinder with 1 cm³ divisions also covers 8 cm³ but has coarser divisions.

  5. Check each choice

    Working

    Answer

    Each instrument’s range includes the value, and its resolution is small compared with the value.

More digits do not make a reading correct. A precise instrument that is not zeroed gives readings that are precise but wrong.

Check your understanding 1 Choose a measuring cylinder

You have three measuring cylinders: 25 cm³ with 0.5 cm³ divisions, 100 cm³ with 1 cm³ divisions and 250 cm³ with 2 cm³ divisions. Which gives the finest reading in one fill for (a) 20 cm³ of water and (b) 180 cm³ of water?

Cylinder for 20 cm³
Cylinder for 180 cm³

Show answer

(a) The 25 cm³ cylinder: it covers 20 cm³, and 0.5 cm³ is finer than 2 cm³. (b) The 250 cm³ cylinder: 180 cm³ does not fit in the 25 cm³ cylinder, and filling it eight times would add a reading error each time.

Choosing a meter range

Many ammeters and voltmeters have more than one range. Suppose an analogue ammeter has a 0–1 A range and a 0–5 A range, each with 50 divisions across its scale.

  • On 0–1 A, one division is 1 ÷ 50 = 0.02 A.
  • On 0–5 A, one division is 5 ÷ 50 = 0.1 A.

To measure a current of about 0.35 A, both ranges cover it, but the 0–1 A range reads it to 0.02 A instead of 0.1 A, so choose 0–1 A. If you do not know the current, start on the largest range and switch down once you see the reading fits. A current larger than the range sends the pointer off the scale and can damage the meter.

Choose a voltmeter range the same way. For the p.d. across a 1.5 V cell, a 0–3 V range covers it; if the 0–3 V and 0–5 V scales have the same number of divisions, the 0–3 V divisions are smaller, so choose 0–3 V.

Check your understanding 2 Select and switch meter ranges

A voltmeter has 0–3 V, 0–6 V and 0–15 V ranges, each with 30 divisions. The p.d. across a component is unknown. Find the size of one division on the 0–3 V and 0–15 V ranges. Which range do you start on? The reading is 2.4 V: what do you do next, and why?

Unit: V
Unit: V
Range to start on
Range after the 2.4 V reading

Show answer

One division is 3 ÷ 30 = 0.1 V on 0–3 V and 15 ÷ 30 = 0.5 V on 0–15 V. Start on 0–15 V, in case the p.d. is large. 2.4 V fits the 0–3 V range, so switch to it: it reads to 0.1 V instead of 0.5 V.

Read the scale correctly

A reading is only as good as the way it is taken. The end of a rule is often worn, so the rod below is lined up with a clear mark instead of the end.

Measuring a rod with a centimetre ruleA rule marked from 0 to 10 cm with millimetre divisions. A rod lies above it with its left end level with the 2.0 cm mark and its right end level with the 7.6 cm mark. Dashed lines drop from each end of the rod to the scale.
The rod does not start at the zero mark, so its length is the difference between the two end readings.

The left end is at 2.0 cm and the right end is at 7.6 cm, so the rod is 7.6 - 2.0 = 5.6 cm long. The rule’s smallest division is 1 mm = 0.1 cm, so the length is recorded to 0.1 cm: 5.6 cm, or 56 mm. Reading only the right end would give 7.6 cm, too long by the 2.0 cm before the rod starts.

Look straight at each mark. Looking from one side makes the mark appear to move against the scale; this is parallax error.

Reading a measuring cylinderA measuring cylinder with labelled marks at 10, 20, 30, 40 and 50 cm³ and four unlabelled marks between each pair, dividing each 10 cm³ into five equal spaces. Water curves up at the walls; the bottom of its curved surface is on the third small mark above 30. An eye on the right looks along a horizontal dashed line level with the bottom of the curve.
Reading a measuring cylinderA measuring cylinder with labelled marks at 10, 20, 30, 40 and 50 cm³ and four unlabelled marks between each pair, dividing each 10 cm³ into five equal spaces. Water curves up at the walls; the bottom of its curved surface is on the third small mark above 30. An eye on the right looks along a horizontal dashed line level with the bottom of the curve.
Read the bottom of the meniscus with your eye level with it; each small division on this cylinder is 2 cm³.

In this cylinder, the labelled marks are 10 cm³ apart with five spaces between each pair, so one small division is 10 ÷ 5 = 2 cm³. Water curves up at the walls. Read the bottom of the curved surface, the meniscus, with your eye level with it: it is three small divisions above 30, so the volume is 30 + 3 × 2 = 36 cm³.

Read a thermometer the same way: keep the bulb in the liquid, stir, and read with your eye level with the end of the liquid thread.

Check the zero

Before you measure, check what the instrument reads when it measures nothing: calipers or micrometer closed, balance empty, spring balance hanging unloaded. If it does not read zero, it has a zero error, and every reading is shifted by that amount.

Reset it to zero if you can. Otherwise, correct each reading:

corrected reading = reading - zero reading

Worked example 2

Correct a zero error

Problem

Digital calipers read 0.03 mm with the jaws closed. On a metal rod they read 8.27 mm. Find the diameter of the rod.

Show full solution
  1. Identify the zero error

    Method

    Read the display with the jaws closed.

    Reason

    Every reading includes this offset.

    Working

    Answer

    Zero error = +0.03 mm, so every reading is 0.03 mm too large.

  2. Correct the reading

    Method

    Subtract the zero error.

    Reason

    Corrected reading = observed reading − zero error.

    Working

    Answer

    8.27 - 0.03 = 8.24 mm

  3. Report it

    Working

    Answer

    In metres, since 1 mm = 10⁻³ m, the diameter is 8.24 × 10⁻³ m. Keep the unit and record digits consistent with the stated instrument’s display step.

A zero error can be negative. A spring balance whose pointer rests at −0.2 N when unloaded reads 3.4 N with a load. The corrected weight is 3.4 - (-0.2) = 3.6 N: subtracting a negative zero reading adds to the reading.

Check your understanding 3 Correct the balance reading

An empty balance reads 0.4 g. With a stone on it, it reads 25.7 g. Find the mass of the stone to the nearest 0.1 g.

Unit: g

Show answer

25.7 - 0.4 = 25.3 g.

Time many oscillations, not one

One complete oscillation of a pendulum is a swing from one side to the other and back. The time for one oscillation is the period.

A digital stopwatch may display 0.01 s steps, but hand-trigger variation is often much larger. Follow the recording convention in your practical task, and use repeats to assess the variation. The last displayed digit may change because of your reaction time.

A pendulum with a period of about 1.8 s shows the problem. A 0.2 s timing error is more than a tenth of a single period. So time many oscillations and divide.

Worked example 3

Period from twenty oscillations

Problem

Twenty oscillations take 36.4 s. Find the period, and the effect of a 0.2 s timing error.

Show full solution
  1. Find the period

    Method

    Divide the total time by the number of oscillations.

    Reason

    One period is the time for one oscillation.

    Working

    Answer

    36.4 ÷ 20 = 1.82 s

  2. Spread the timing error

    Working

    Answer

    The same 0.2 s timing error is spread over 20 periods: it becomes 0.2 ÷ 20 = 0.01 s in the period.

Start the stopwatch as the bob passes the centre in a stated direction, counting that pass as “zero”. Count one complete oscillation each time it returns through the centre in the same direction. A pass in the opposite direction occurs only halfway through the cycle. The bob moves fastest at the centre, making the crossing easier to judge.

Check your understanding 4 Compare timing methods

A student can either time one pendulum oscillation, about 1.5 s, or time ten oscillations, about 15 s. The total timing error is about 0.2 s in either method. Compare the error in the period found by each method. Would a stopwatch with more display digits remove this error?

Unit: s
Unit: s
Would extra display digits remove this error?

Show answer

Timing one oscillation gives an error of about 0.2 s in the period. Timing ten and dividing by ten gives an error of about 0.2 ÷ 10 = 0.02 s in the period, ten times smaller. More display digits would not remove the error because it comes from starting and stopping the stopwatch by hand.

Repeat readings and take the mean

Readings of the same quantity differ a little each time: your timing varies, and a wire is never perfectly round. Take each reading at least three times and use the mean:

mean = (sum of the readings)/(number of readings)

Worked example 4

Mean of repeated timings

Problem

A student times 20 oscillations three times: 36.2 s, 36.6 s and 36.5 s. Find the period.

Show full solution
  1. Find the mean time

    Method

    Add the readings and divide by 3.

    Reason

    Repeating and averaging reduces the effect of random timing errors.

    Working

    Answer

    (36.2 + 36.6 + 36.5) ÷ 3 = 36.4 s, to 0.1 s like the readings

  2. Find the period

    Working

    Answer

    36.4 ÷ 20 = 1.82 s

Before averaging, compare the readings. If a fourth reading were 34.6 s, it would be about one period short of the others, which suggests a miscounted oscillation. Check the count and repeat the measurement. Keep the original reading and record the investigation; do not exclude it merely because it differs. Exclude a reading only when the investigation identifies an error, such as an observed miscount.

For a wire, measure the diameter at several places along it and in two directions at right angles. Readings of 0.452 mm, 0.448 mm and 0.455 mm give a mean of 1.355 ÷ 3 = 0.452 mm.

Measure many to find one

When one item is too small compared with the instrument’s scale or display step, measure many comparable items together, then divide to estimate a typical item.

Worked example 5

The thickness of one sheet of paper

Problem

Find the thickness of one sheet of paper with a metre rule.

Show full solution
  1. Decide whether to measure directly

    Method

    Compare the sheet with the rule’s smallest division.

    Reason

    One sheet is much thinner than 1 mm, so it cannot be measured directly.

    Working

    Answer

    Measure a stack instead

  2. Measure a stack

    Method

    Press 200 sheets flat and measure them.

    Reason

    A thick stack is many divisions long, so the reading is precise.

    Working

    Answer

    2.0 cm = 20 mm

  3. Divide

    Working

    Answer

    One sheet = 20 ÷ 200 = 0.10 mm = 1.0 × 10⁻⁴ m, which matches the order of magnitude of a sheet of paper.

The same idea works on a balance: 50 identical paper clips with a mass of 23.5 g have a mass of 23.5 ÷ 50 = 0.47 g each.

Some quantities are worked out from other measurements. The volume of an irregular stone is found by displacement: water in a measuring cylinder reads 50 cm³, and 62 cm³ with the stone fully under water, so the stone’s volume is 62 - 50 = 12 cm³.

Check your understanding 5 Measure a thin sheet

You need the thickness of a thin aluminium sheet, about 0.2 mm, as reliably as you can. Choose the instrument, justify it by range and precision, and say how you would take the readings.

Show answer

Use a digital micrometer. Its range, 0–25 mm, covers 0.2 mm, and its resolution, 0.001 mm, is small compared with 0.2 mm. The stated calipers’ 0.01 mm step would be a twentieth of the thickness, and a metre rule’s 1 mm division is larger than the sheet.

Close the micrometer and check that it reads zero; correct any zero error. Close it gently on the sheet, then repeat at several places across the sheet. Take the mean.

Check your understanding 6 Weigh a counted sample

A grain of rice has a mass of about 0.02 g, and your balance reads to 0.1 g. Explain why weighing one grain is not good enough, and describe a better method.

Show answer

One grain, about 0.02 g, is smaller than the balance’s 0.1 g display step, so a correctly zeroed display would round it to 0.0 g. Zero the balance, weigh a counted sample of many grains, such as 100, and divide. If 100 grains have a mass of 2.1 g, the mean mass per grain is 2.1 ÷ 100 = 0.021 g; real grains need not all have that mass.

Check your understanding 7 Count complete oscillations

A pendulum crosses the centre travelling right when the timer starts. The timer stops 18.0 s later at the eleventh rightward centre crossing, including the starting crossing. Find the number of complete oscillations and the period. Another learner counts every centre crossing, including leftward ones, as a full oscillation. Choose what the next leftward crossing represents.

Unit: cycles
Show the steps Optional working
Unit: cycles
Unit: s
What does the next leftward crossing mark?

Show answer

The first rightward crossing starts the first interval; the eleventh completes ten full cycles. Period = 18.0 ÷ 10 = 1.80 s. Between two rightward crossings there is a leftward crossing, only half a cycle later. Counting every centre crossing as a full oscillation doubles the count and underestimates the period. Start and finish at the same position and direction; averaging does not correct a wrong cycle count.

Common mistakes

  • Choosing the instrument with the most digits. Check that the range covers the value first, then compare its scale or display step with the value and differences to detect.
  • Expecting a more precise instrument to fix a zero error. Check the zero, then reset it or subtract the zero reading.
  • Adding a positive zero error. The instrument already reads too high: corrected reading = reading − zero reading.
  • Treating the stopwatch’s last digit as proof of accuracy. Time several complete oscillations and repeat; the hand-trigger variation can be larger than the display step.
  • Forgetting to divide by the number of oscillations or items. That gives the total, not one.
  • Reading the top of the meniscus, or looking from an angle. Read the bottom of the meniscus with your eye level with it.
  • Reading only the far end of an object. Subtract the start reading from the end reading.
  • Starting a meter on its smallest range. Start on the largest range, then switch down.
  • Deleting an unusual reading without a justified, recorded reason. Investigate and repeat; keep the original result.

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