Physics · Chapter 2: Physical Quantities and Measurement

Vernier Callipers

Parts of vernier callipers, least count 0.01 cm, taking a reading, and positive and negative zero error with corrected readings.

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01Concept

A metre rule cannot measure less than 1 mm. To measure the diameter of a pencil or a coin more precisely, we use vernier callipers. It was invented by the French mathematician Pierre Vernier in 1631. Its least count is 0.1 mm (0.01 cm), ten times smaller than a metre rule.

Vernier callipers. The outside jaws measure outer sizes, the inside jaws measure inside diameters and the depth rod measures depths.

Parts

  • Main scale: a fixed scale in centimetres and millimetres.
  • Vernier scale: a small sliding scale. 10 vernier divisions equal 9 mm on the main scale, so each vernier division is 0.9 mm.
  • Outside jaws: for the length or outer diameter of an object.
  • Inside jaws: for the inner diameter of a pipe or beaker.
  • Depth rod: for the depth of a hole or a beaker.
  • Lock screw: holds the sliding jaw in place while you read.

Least count

The difference between one main scale division (1 mm) and one vernier division (0.9 mm) is the least count:

Least count = 1 mm − 0.9 mm = 0.1 mm = 0.01 cm, or least count = smallest main scale reading ÷ number of vernier divisions = 1 mm ÷ 10

Taking a reading

  1. Close the jaws gently on the object.
  2. Read the main scale reading (MSR): the main scale mark just before the vernier zero. 2.4 cm
  3. Find the vernier division that lines up exactly with any main scale mark. The 6th division.
  4. Vernier scale reading (VSR) = coinciding division × least count = 6 × 0.01 cm = 0.06 cm.
  5. Total reading = MSR + VSR = 2.4 + 0.06 = 2.46 cm

Zero error

When the jaws are closed, the zero of the vernier should line up with the zero of the main scale. If it does not, the instrument has a zero error.

Positive zero errorNegative zero error
Position of vernier zero (jaws closed)To the right of the main scale zeroTo the left of the main scale zero
EffectEvery reading is too largeEvery reading is too small
Example+0.03 cm−0.04 cm

Corrected reading = observed reading − zero error. Observed 2.46 cm, zero error +0.03 cm → corrected = 2.43 cm. With a zero error of −0.04 cm → 2.46 − (−0.04) = 2.50 cm.

02Virtual Lab

Try this topic for yourself in the virtual lab. It opens on its own page with steps, tasks and questions.

03Key Formulas

Least count of vernier callipers

LC=one main scale divisionnumber of vernier divisions=1 mm10=0.1 mm\text{LC} = \frac{\text{one main scale division}}{\text{number of vernier divisions}} = \frac{1\ \text{mm}}{10} = 0.1\ \text{mm}

Vernier reading

reading=MSR+(n×LC)\text{reading} = \text{MSR} + (n \times \text{LC})
  • MSR\text{MSR} = main scale reading
  • nn = vernier division that coincides

Corrected reading

corrected=observed−zero error\text{corrected} = \text{observed} - \text{zero error}

Keep the sign of the zero error.

04Real World Examples

Mechanics' workshops

Motor mechanics use vernier callipers (often digital ones) to check the thickness of brake discs and the size of nuts and bolts.

Jewellers

Goldsmiths measure the size of rings and the thickness of bangles with small vernier callipers.

Digital callipers

A digital calliper shows the reading on a screen, but inside it works on the same idea and still has a zero button to remove zero error.

05Practice Questions

Try to answer in your own words first, then open the model answer.

Short questions

Q1.What is the least count of vernier callipers? How is it found?Show answer

Answer

Least count = 1 mm ÷ 10 = 0.1 mm = 0.01 cm. It is the difference between one main scale division (1 mm) and one vernier division (0.9 mm).

Q2.What is zero error? When is it positive?Show answer

Answer

Zero error is a reading shown by an instrument when it should read zero. In vernier callipers it is positive when the vernier zero lies to the right of the main scale zero with the jaws closed.

Q3.The main scale reading is 3.2 cm and the 7th vernier division coincides. Find the reading.Show answer

Answer

  • VSR = 7 × 0.01 = 0.07 cm
  • Reading = MSR + VSR = 3.2 + 0.07
  • = 3.27 cm
Q4.A vernier calliper has a zero error of −0.02 cm. The observed reading is 1.84 cm. Find the corrected reading.Show answer

Answer

  • Corrected = observed − zero error
  • = 1.84 − (−0.02)
  • = 1.86 cm
Q5.Name the three uses of vernier callipers.Show answer

Answer

  • Outside jaws: outer length or diameter, e.g. a coin.
  • Inside jaws: inner diameter, e.g. a beaker.
  • Depth rod: depth, e.g. of a test tube.

Long questions

Q6.Describe vernier callipers. How would you find the diameter of a cylinder with it? Include zero error.Show answer

Answer

  1. Find the least count: 1 mm ÷ 10 = 0.01 cm.
  2. Close the jaws and check the zero error; note its sign.
  3. Place the cylinder between the outside jaws and close them gently. Tighten the lock screw.
  4. Read the main scale reading just before the vernier zero, e.g. 2.4 cm.
  5. Find the coinciding vernier division, e.g. 6. VSR = 6 × 0.01 = 0.06 cm.
  6. Observed reading = 2.4 + 0.06 = 2.46 cm.
  7. Corrected reading = observed − zero error. With +0.03 cm: 2.43 cm.
  8. Repeat at different places on the cylinder and take the mean.

06Quiz

Question 1 of 8
The least count of vernier callipers is: