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WASSCE Physics · Measurement and Errors

Measuring length: metre rule, vernier calipers and micrometer screw gauge

By the WASSCE Tutor team

Updated 11 November 2026 · 12 min read · Prepared against the WASSCE syllabus.

A metre rule can tell you a pencil is 17.5 cm long, but it cannot measure the thickness of a strand of wire. Choosing the right instrument, and reading it correctly, earns easy marks in both the theory and the practical paper. This lesson covers the three length instruments WASSCE tests most: the metre rule, vernier calipers and the micrometer screw gauge.

Learning objectives

By the end of this lesson you should be able to:

  • choose a suitable instrument for a given length, from a desk to a wire
  • explain least count and work it out for a vernier and a micrometer
  • take and record a reading from vernier calipers and a micrometer screw gauge
  • correct a reading for positive and negative zero error

Choosing the right instrument

Every measurement has three parts: a quantity, a number and a unit. "Length = 0.842 m" names the quantity (length), gives the value (0.842) and the unit (metre). No instrument is perfectly exact, so the instrument must suit the size of what you are measuring.

Definition
  • The range of an instrument is the largest and smallest values it can measure.
  • The least count is the smallest change it can show: the value of its smallest scale division (or of one vernier or thimble division).
InstrumentTypical useLeast count
Measuring tapelong or curved distances1 mm (0.1 cm)
Metre rulelengths on a bench1 mm (0.1 cm)
Vernier calipersexternal and internal diameters, depth0.1 mm (0.01 cm)
Micrometer screw gaugethin wires, sheets, small balls0.01 mm (0.001 cm)

The smaller the least count, the finer the reading. But a fine scale does not make a reading accurate if the instrument is damaged or badly zeroed.

The metre rule

Lay the rule along the object, not at an angle. If the object does not start at the zero mark (the end of an old rule is often worn), read both ends and subtract.

Worked example

A rod starts at the 1.4 cm mark and ends at the 18.9 cm mark. How long is it?

L=final reading−initial reading=18.9−1.4=17.5 cmL = \text{final reading} - \text{initial reading} = 18.9 - 1.4 = 17.5\ \text{cm}
L=final reading−initial reading=18.9−1.4=17.5 cm\begin{aligned}L &= \text{final reading} \\ &\quad - \text{initial reading} \\ &= 18.9 - 1.4 \\ &= 17.5\ \text{cm}\end{aligned}

The rod is 17.5 cm long, not 18.9 cm.

Looking at a scale from one side makes the mark seem to shift against the scale. This is parallax error. Avoid it by placing your eye directly above the mark you are reading.

Vernier calipers

Vernier calipers have a fixed main scale on a steel beam and a sliding vernier scale on the moving jaw. The large lower jaws grip the outside of an object; the small upper jaws fit inside a hole or tube; the thin rod at the end measures depth.

Vernier calipers holding a block between the outside jaws, with the main scale on the beam, the vernier scale on the sliding jaw, inside jaws on top, a locking screw and a depth rod

Figure 1. The parts of vernier calipers. WASSCE Tutor diagram.

Least count of a vernier

On the usual school vernier, 10 vernier divisions have the same length as 9 main-scale divisions (9 mm). So one vernier division is 0.9 mm, and:

least count=1 MSD−1 VSD=1.0 mm−0.9 mm=0.1 mm\text{least count} = 1\ \text{MSD} - 1\ \text{VSD} = 1.0\ \text{mm} - 0.9\ \text{mm} = 0.1\ \text{mm}
least count=1 MSD−1 VSD=1.0 mm−0.9 mm=0.1 mm\begin{aligned}&\text{least count} \\ &= 1\ \text{MSD} - 1\ \text{VSD} \\ &= 1.0\ \text{mm} - 0.9\ \text{mm} \\ &= 0.1\ \text{mm}\end{aligned}

Taking a reading

  1. Read the main scale at the mark just before (to the left of) the vernier zero. This gives whole millimetres.
  2. Find the one vernier mark that lines up exactly with any main-scale mark. Call its number nn.
  3. Reading == main-scale reading + n×+\ n \times least count.

Close-up of a vernier scale: the vernier zero lies just past 24 mm on the main scale and the sixth vernier mark lines up with a main-scale mark, giving 24.6 mm

Figure 2. Reading vernier calipers. WASSCE Tutor diagram.

Worked example

In Figure 2, the vernier zero lies just past 24 mm and the 6th vernier mark lines up with a main-scale mark. Find the reading.

reading=24 mm+6×0.1 mm=24.6 mm=2.46 cm\text{reading} = 24\ \text{mm} + 6 \times 0.1\ \text{mm} = 24.6\ \text{mm} = 2.46\ \text{cm}
reading=24 mm+6×0.1 mm=24.6 mm=2.46 cm\begin{aligned}&\text{reading} \\ &= 24\ \text{mm} + 6 \times 0.1\ \text{mm} \\ &= 24.6\ \text{mm} \\ &= 2.46\ \text{cm}\end{aligned}
Common mistake

Students often read the main-scale mark after the vernier zero (25 mm here) or add 6 mm instead of 6 × 0.1 mm. Always take the last main-scale mark before the vernier zero, and multiply the coinciding division by the least count.

Micrometer screw gauge

A micrometer measures very small lengths, such as the diameter of a wire or the thickness of a sheet of paper. The object sits between the fixed anvil and the moving spindle. Turning the thimble screws the spindle in or out.

Micrometer screw gauge with a C-shaped frame, fixed anvil, object, spindle, sleeve carrying the main scale, thimble carrying the circular scale, and a ratchet at the end

Figure 3. The parts of a micrometer screw gauge. WASSCE Tutor diagram.

Least count of a micrometer

Definition

The pitch of the screw is the distance the spindle moves in one complete turn of the thimble. For most school micrometers the pitch is 0.5 mm and the thimble has 50 divisions.

least count=pitchnumber of thimble divisions=0.5 mm50=0.01 mm\text{least count} = \dfrac{\text{pitch}}{\text{number of thimble divisions}} = \dfrac{0.5\ \text{mm}}{50} = 0.01\ \text{mm}
least count=pitchnumber of thimble divisions=0.5 mm50=0.01 mm\begin{aligned}&\text{least count} \\ &= \dfrac{\text{pitch}}{\text{number of thimble divisions}} \\ &= \dfrac{0.5\ \text{mm}}{50} \\ &= 0.01\ \text{mm}\end{aligned}

Taking a reading

  1. Close the spindle on the object by turning the ratchet, not the thimble, until it clicks. This applies the same gentle force every time and does not squash the object.
  2. Read the sleeve: whole millimetres from the marks above the datum line, plus 0.5 mm if a half-millimetre mark below the line is also showing.
  3. Read the thimble division that lies on the datum line. Call it nn.
  4. Reading == sleeve reading + n×0.01+\ n \times 0.01 mm.

Close-up of a micrometer: the sleeve shows 5 mm and the half-millimetre mark after it, and the thimble division 28 lies on the datum line, giving 5.78 mm

Figure 4. Reading a micrometer screw gauge. WASSCE Tutor diagram.

Worked example

In Figure 4, the sleeve shows 5 mm and the half-millimetre mark after it, and division 28 of the thimble is on the datum line. Find the diameter of the wire.

reading=5.5 mm+28×0.01 mm=5.5+0.28=5.78 mm\text{reading} = 5.5\ \text{mm} + 28 \times 0.01\ \text{mm} = 5.5 + 0.28 = 5.78\ \text{mm}
reading=5.5 mm+28×0.01 mm=5.5+0.28=5.78 mm\begin{aligned}&\text{reading} \\ &= 5.5\ \text{mm} + 28 \times 0.01\ \text{mm} \\ &= 5.5 + 0.28 \\ &= 5.78\ \text{mm}\end{aligned}
Common mistake

The commonest micrometer error is missing the half-millimetre mark. If it is showing, the sleeve reading is 5.5 mm, not 5 mm, and an answer of 5.28 mm would lose the mark.

Zero error and its correction

An instrument has a zero error if it does not read zero when it should: when the jaws of the calipers are closed, or when the spindle of the micrometer touches the anvil.

  • Positive zero error: the instrument reads above zero when closed (for example +0.2+0.2 mm). Every reading is too large.
  • Negative zero error: the zero mark has gone past the datum (for example −0.03-0.03 mm). Every reading is too small.
corrected reading=observed reading−zero error\text{corrected reading} = \text{observed reading} - \text{zero error}
corrected reading=observed reading−zero error\begin{aligned}&\text{corrected reading} \\ &= \text{observed reading} \\ &\quad - \text{zero error}\end{aligned}
Worked example

(a) Calipers read +0.2 mm when closed and 16.7 mm on an object. (b) A micrometer has a zero error of −0.03 mm and reads 2.46 mm on a wire. Find the corrected readings.

(a) 16.7−(+0.2)=16.516.7 - (+0.2) = 16.5 mm

(b) 2.46−(−0.03)=2.46+0.03=2.492.46 - (-0.03) = 2.46 + 0.03 = 2.49 mm

Always write the sign of the zero error before you subtract. Subtracting a negative zero error means adding.

In Sierra Leone

In fitting shops and garages across Freetown, Bo and Makeni, mechanics check the diameter of bolts, pistons and pipes with vernier calipers, because a part that is 0.5 mm too wide will not fit. The same reading method you learn here is used on the job every day.

Which instrument for which job?

To measureUse
the length of a laboratory benchmetre rule or tape
the internal diameter of a test tubevernier calipers (inside jaws)
the depth of a beakervernier calipers (depth rod)
the diameter of a pendulum bobvernier calipers (outside jaws)
the diameter of a thin wire or the thickness of a sheet of papermicrometer screw gauge

Key points

  • Choose an instrument whose least count suits the size of the measurement.
  • Vernier: least count 0.1 mm; reading = main scale (mark before the vernier zero) + n×0.1n \times 0.1 mm.
  • Micrometer: least count = pitch ÷ thimble divisions = 0.01 mm; reading = sleeve (include the half-mm mark) + n×0.01n \times 0.01 mm.
  • Corrected reading = observed reading − zero error, keeping the sign.
  • Avoid parallax by viewing scales straight on; close a micrometer with the ratchet.

WASSCE examination focus

WAEC tests this

Expect to name the instrument for a given measurement, work out the least count from scale information, take a reading from a drawn vernier or micrometer scale, and apply a zero correction with the correct sign. In the practical paper, record the instrument reading to its least count (for example 2.46 cm from calipers, 0.58 mm from a micrometer) and state precautions such as "I used the ratchet to avoid over-tightening" or "I checked the zero error before taking readings".

Quick check

  1. Ten vernier divisions equal nine main-scale divisions of 1 mm. What is the least count?
  2. A vernier reads 31 mm on the main scale and the 4th vernier mark coincides. What is the reading?
  3. A micrometer sleeve shows 3.5 mm and division 36 is on the datum line. What is the reading?
  4. Calipers with a zero error of +0.04 cm read 8.76 cm. What is the corrected reading?
  5. Which instrument would you use to measure the diameter of a thin copper wire, and why?

Answers: 1. 1.0−0.9=0.11.0 - 0.9 = 0.1 mm. 2. 31+4×0.1=31.431 + 4 \times 0.1 = 31.4 mm. 3. 3.5+36×0.01=3.863.5 + 36 \times 0.01 = 3.86 mm. 4. 8.76−0.04=8.728.76 - 0.04 = 8.72 cm. 5. A micrometer screw gauge: its least count (0.01 mm) is small enough for a diameter of a fraction of a millimetre, and its ratchet applies a steady force.

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