Practice On Reading A Vernier Caliper
Key idea: Vernier caliper reading practice (no zero error): 9 diagram questions with worked answers in cm to help O Level Physics students.
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The core idea
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Learning objectives
- Read an analogue vernier caliper scale and correct the reading for zero error
1. Definition
A. What you are practising
Practise reading a vernier caliper (no zero error). For each question, give the observed reading in cm to 0.01 cm.
If you need a refresher, use: How To Use Calipers (Digital + Vernier).
2. Key Ideas
- Main scale reading: the value immediately left of the vernier zero.
- Vernier reading: the aligned vernier division × 0.01 cm.
- Observed reading = main scale + vernier reading.
3. Detailed Explanations
A. Quick workflow (exam-friendly)
- Write the main scale reading (to 0.1 cm).
- Write the vernier reading (aligned division × 0.01 cm).
- Add them to get the observed reading (to 0.01 cm).
4. Common Mistakes
A. Reading mistakes
- Using the main-scale value to the right of the vernier zero.
- Picking a line that “almost” aligns (choose the best alignment).
- Writing the unit wrongly (mm vs cm).
5. Exam Tips
A. What markers like to see
- Show the two parts clearly: “main scale” + “vernier”.
- Keep units in every line, then give a final answer in cm.
5A. Interactive Trainer
Use the simulation to generate random vernier readings, then use the fixed questions below to lock in exam method.
Simulation Trainer: Vernier Caliper Reading
Train the no-zero-error vernier workflow: main scale, aligned division, final reading in cm.
- Main Scale Reading
- Vernier Alignment
- Instrument Precision
6. Worked Examples
Modelled example 1
Question 1 — modelled reading
Problem
Study the worked solution
Read the main scale
Method
The main-scale reading immediately left of vernier zero is 8.6 cm.Reason
The value to the right has not yet been passed.Working
xₘₐᵢₙ = 8.6 cmRead the vernier
Method
Division 2 aligns, giving 0.02 cm.Reason
Each vernier division represents 0.01 cm.Working
xᵥₑᵣₙᵢₑᵣ = 2(0.01) = 0.02 cmAdd the readings
Method
The observed reading is 8.62 cm.Reason
The vernier part refines the main-scale reading.Working
x = 8.6 + 0.02 = 8.62 cm
Guided practice 2
Question 2 — zero main-scale contribution
Problem
Try this before viewing the solution
Hints
Hint 1: separate the two scale parts
View solution step by step
Read both parts
Method
The main scale is 0.0 cm and division 6 contributes 0.06 cm.Reason
Aligned vernier division n contributes n × 0.01 cm.Working
0.0 + 6(0.01)State the reading
Method
The reading is 0.06 cm.Reason
The zero main-scale contribution does not remove the vernier contribution.Working
x = 0.0 + 0.06 = 0.06 cm
Guided practice 3
Question 3 — combine both scales
Problem
Try this before viewing the solution
Hints
Hint 1: convert division 3
View solution step by step
Read both parts
Method
The main scale is 6.4 cm and the vernier contributes 0.03 cm.Reason
Division 3 is the aligned vernier mark.Working
xᵥₑᵣₙᵢₑᵣ = 3(0.01) = 0.03 cmAdd
Method
The reading is 6.43 cm.Reason
Observed reading equals main plus vernier.Working
x = 6.4 + 0.03 = 6.43 cm
Common misconception 4
Question 4 — place-value trap
Learner claim
Try this before viewing the solution
View solution step by step
Diagnose the place value
Method
Division 6 means 0.06 cm, not 0.60 cm.Reason
The least count is 0.01 cm per vernier division.Working
6(0.01) = 0.06 cmCorrect the reading
Method
The observed reading is 1.06 cm.Reason
Add the vernier contribution to the 1.0 cm main scale.Working
x = 1.0 + 0.06 = 1.06 cm
Common misconception 5
Question 5 — choose the main-scale mark on the left
Learner claim
Try this before viewing the solution
View solution step by step
Correct the main scale
Method
The main-scale reading is 5.3 cm.Reason
The vernier zero has passed 5.3 cm but not 5.4 cm.Working
xₘₐᵢₙ = 5.3 cmAdd the vernier
Method
The reading is 5.31 cm.Reason
Aligned division 1 adds 0.01 cm.Working
x = 5.3 + 1(0.01) = 5.31 cm
Examiner practice 6
Question 6 — examiner working
Examination question
Try this before viewing the solution
View solution step by step
Main scale
1 markMethod
3.8 cm.Reason
This is immediately left of vernier zero.Working
xₘₐᵢₙ = 3.8 cmVernier
1 markMethod
0.03 cm.Reason
Division 3 aligns.Working
3(0.01) = 0.03 cmTotal
1 markMethod
3.83 cm.Reason
Add the two scale contributions.Working
3.8 + 0.03 = 3.83 cm
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark the main scale, vernier contribution and final reading.
Examiner practice 7
Question 7 — high aligned division
Examination question
Try this before viewing the solution
View solution step by step
Resolve the scales
2 marksMethod
The parts are 4.2 cm and 0.07 cm.Reason
Division 7 aligns and each division is 0.01 cm.Working
xᵥₑᵣₙᵢₑᵣ = 7(0.01) = 0.07 cmAdd
1 markMethod
The reading is 4.27 cm.Reason
Observed reading is main plus vernier.Working
x = 4.2 + 0.07 = 4.27 cm
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark both scale parts and their sum.
Examiner practice 8
Question 8 — whole-centimetre main scale
Examination question
Try this before viewing the solution
View solution step by step
Resolve the scales
2 marksMethod
The parts are 7.0 cm and 0.05 cm.Reason
Division 5 aligns.Working
xᵥₑᵣₙᵢₑᵣ = 5(0.01) = 0.05 cmAdd
1 markMethod
The reading is 7.05 cm.Reason
The two contributions have different decimal places.Working
x = 7.0 + 0.05 = 7.05 cm
Self-mark with the mark scheme
Compare your response with each mark point. Select a point only when your response contains that evidence.
Self-mark both scale parts and the final precision.
Challenge 9
Question 9 — near-zero reading and unit transfer
Independent transfer
Try this before viewing the solution
Hints
Hint 1: finish the instrument reading first
View solution step by step
Read the scale
Method
The observed reading is 0.24 cm.Reason
The main scale is 0.2 cm and division 4 adds 0.04 cm.Working
x = 0.2 + 4(0.01) = 0.24 cmConvert the representation
Method
The same length is 2.4 mm.Reason
1 cm = 10 mm.Working
0.24 × 10 = 2.4 mm
7. Mind Stretchers
Mind stretcher 1: Convert your answer to SI unitExtension
Convert 8.62 cm to metres.
Show Answer
8.62 cm = 0.0862 m.
Mind stretcher 2: Parallax checkExtension
Why must you read the vernier scale at eye level?
Show Answer
To avoid parallax error: reading at an angle shifts the apparent alignment and gives the wrong vernier division.
Continue with the next resource in this course.
Course and syllabus information
- Course
- G3 Physics topic extensions
- Syllabus scope
- Beyond the syllabus
- Edition
- G3 Physics topic extensions