Vernier Callipers: two scales that read a tenth of a millimetre

The problem

A pencil is about 7.0 × 10⁻³ m thick, 7 mm. A metre rule stops at whole millimetres. To see the tenths, the French scientist Pierre Vernier added a second, sliding scale in 1631.

The main scale has 1 mm divisions. The Vernier scale is 9 mm long and split into 10 equal parts, so each part is 0.9 mm. The difference, 1 mm − 0.9 mm = 0.1 mm, is the least count.

To read: take the main scale mark just before the Vernier zero. Then find the Vernier division that lines up with a main scale mark. Length = main scale reading + least count × Vernier reading.

Close the jaws first. If the two zeros do not coincide, there is a zero error. Vernier zero to the right: the instrument reads more, so subtract it. To the left: it reads less, so add it.

Least count = 1 main scale division − 1 Vernier division = 0.1 mm. Length = main scale reading + least count × Vernier reading, then correct for any zero error.

Key terms

Main scale

The fixed scale; it gives the whole millimetres.

Vernier scale

The sliding scale; the division that lines up gives the tenths.

Least count (Vernier)

One main scale division minus one Vernier division.

Zero error

The reading shown when the jaws are closed and the zeros do not coincide.

Short questions with model answers

1. Q1. MCQ 1.11 (p. 24): ten Vernier divisions equal nine main scale divisions, and one main scale division is half a millimetre. Find the least count.

   • least count = 1 main scale division ÷ 10 Vernier divisions
   • 0.5 mm ÷ 10 = 0.05 mm

   0.05 mm

2. Q2. The book's zero error: jaws closed, Vernier zero to the left, division 3 coincides. Find the zero error and the correction.

   • (10 − 3) × 0.1 mm = 0.7 mm
   • Add 0.7 mm to every reading.

   Zero error 0.7 mm; add it.

3. Q3. CRQ 1.10 (p. 26): which part measures (i) the internal diameter of a test tube, (ii) the depth of a beaker?

   • (i) Inside jaws C and D.
   • (ii) The tail, or depth gauge.

   Inside jaws; tail (depth gauge).

Common mistakes

• ✗ Adding 4.3 cm + 0.1 × 4 = 4.7 cm, mixing centimetres and millimetres.

  ✓ Put everything in one unit: 0.1 mm = 0.01 cm, so 4.3 + 0.01 × 4 = 4.34 cm.

• ✗ Always subtracting the zero error.

  ✓ Right of zero: it reads more, subtract. Left of zero: it reads less, add.

• ✗ Taking the main scale mark after the Vernier zero.

  ✓ Take the mark just before the Vernier zero; the Vernier adds the part beyond it.

MCQs

1. 1. The least count of the Vernier Callipers in your book is:

   1. (a) 1 mm
   2. (b) 0.1 mm
   3. (c) 0.01 mm
   4. (d) 0.9 mm
   Show answer

   (b) 1 mm − 0.9 mm = 0.1 mm.

2. 2. The depth of a beaker is measured with the:

   1. (a) Outside jaws
   2. (b) Inside jaws
   3. (c) Main scale alone
   4. (d) Tail (depth gauge)
   Show answer

   (d) The tail measures the depth of hollow objects (p. 13).

3. 3. With the jaws closed, the Vernier zero is to the right of the main zero. The zero error is:

   1. (a) Subtracted from the reading
   2. (b) Added to the reading
   3. (c) Ignored
   4. (d) Multiplied by 10
   Show answer

   (a) It reads slightly more than the true length, so subtract (p. 14).

4. 4. MCQ 1.9 (p. 24): four students measure a cylinder with Vernier Callipers. Which reading is correctly recorded?

   1. (a) 3.4 cm
   2. (b) 3.475 cm
   3. (c) 3.47 cm
   4. (d) 3.5 cm
   Show answer

   (c) The least count is 0.01 cm, so the reading has two decimal places in cm.

5. 5. Main scale reading 4.3 cm, Vernier division 4. The length is:

   1. (a) 4.7 cm
   2. (b) 4.34 cm
   3. (c) 8.3 cm
   4. (d) 4.304 cm
   Show answer

   (b) 4.3 + 0.01 × 4 = 4.34 cm (Fig. 1.3).

Quick revision

• Least count = 1 mm − 0.9 mm = 0.1 mm (0.01 cm).
• Length = main scale reading + least count × Vernier reading: 4.34 cm in Fig. 1.3.
• Zero error: subtract if the Vernier zero is right, add if it is left.