Vernier caliper reading practice

Read main and vernier scales, correct signed zero errors and find resolution.

Practise

Negative zero error

The signed zero error (indication when the jaws close) is −0.40 mm. Find the corrected length in mm.

Main divisions are 1 mm; 10 vernier divisions span 9 mm. Read the aligned index from 0 to 9. At index 0, the final endpoint also aligns. This is a scale close-up.

Read the main scale above and the vernier below. Scroll the fine scale sideways on a small screen.

Diagram text alternative

Main-scale mark immediately left of vernier zero: 10 mm. Aligned vernier division: 5. 10 vernier divisions span 9 mm.

Larger diagram

Hint

Corrected length = indicated reading − signed zero error.

Worked solution

Main scale = 10 mm.
Resolution = 1 mm − 9/10 mm = 0.10 mm.
Vernier contribution = 5 × 0.10 = 0.50 mm.
Indicated reading = 10.50 mm.
10.50 − (-0.40) = 10.90 mm.

Another question

Question type: Mixed practice

Report a problem

Read the main scale and vernier scale

For a vernier with resolution 0.1 mm, suppose the main-scale reading just left of the vernier zero is 24 mm and the sixth vernier line aligns. Reading = 24 + 6 × 0.1 = 24.6 mm. Use the instrument’s stated resolution; different verniers have different increments.

The main-scale reading is the last mark before the vernier zero. The aligning line contributes a fraction of a main-scale division, not six extra millimetres. Apply a zero-error correction only when one is supplied or established.

Compare with a micrometer reading

Read the main scale first

Use the main-scale mark immediately to the left of vernier zero. Add the number of the aligned vernier mark multiplied by the resolution. The main scale here has 1 mm divisions. Use an aligned index from 0 through N − 1. When index 0 aligns, the final endpoint N also aligns; that endpoint repeats the zero position.

Find the resolution

If N vernier divisions span N − 1 mm, each vernier division is (N − 1)/N mm. The difference from a 1 mm main division is 1/N mm: 0.1 mm for N = 10 or 0.05 mm for N = 20.

Correct the zero

Corrected length = indicated length − signed zero error. Subtract a positive error; add the magnitude of a negative one. The zero error is supplied as the indication when the jaws close. Scale marks align exactly in these exercises.

A-level relevance

AQA 7408: 7.1 AT e (selected); OCR A H556: 1.2.2(e) (selected). Tasks cover the selected skills listed above.

AQA specification · OCR A specification · AQA practical handbook · OCR practical skills handbook