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Vernier caliper and micrometer readings

Short answer

(a) Vernier principle: n VSD = (n-1) MSD, LC = MSD - VSD = MSD/n. (b) LC = 0.02 mm; observed =34.54 mm, zero error +0.06 mm, corrected diameter =34.48 mm. (c) LC = 0.01 mm; observed =12.77 mm, zero error -0.02 mm, corrected =12.79 mm.

(a) Explain the principle of a vernier caliper and obtain the expression for its least count. (b) A vernier caliper has 1 mm main scale divisions, and 50 vernier divisions coincide with 49 mm of the main scale. When measuring a shaft, the zero of the vernier lies just beyond the 34 mm mark and the 27th vernier division coincides with a main scale line. With the jaws closed, the vernier zero is 3 vernier divisions ahead of the main scale zero. Find the least count and the corrected diameter. (c) A micrometer has a pitch 0.5 mm and 50 thimble divisions. Its reading on a pin shows 12.5 mm on the sleeve and the 27th thimble division on the datum line; the zero error is -0.02 mm. Find the corrected reading.

(a) Principle and least count

A vernier scale slides along the main scale. Its nn divisions equal (n−1)(n-1) divisions of the main scale, so each vernier division is slightly shorter than a main scale division. The difference of one main division and one vernier division is the least count (LC), the smallest length that can be read:

LC=1 MSD−1 VSD=1 MSDn\text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{1\ \text{MSD}}{n}

The reading is: main scale reading (just left of the vernier zero) + (coinciding vernier division ×\times LC).

(b) Vernier caliper

1 VSD =49/50=0.98= 49/50 = 0.98 mm, so

LC=1−0.98=0.02 mm\text{LC} = 1 - 0.98 = 0.02\ \text{mm}

Observed reading =34+27×0.02=34.54= 34 + 27 \times 0.02 = 34.54 mm.

Zero error is positive: +3×0.02=+0.06+3 \times 0.02 = +0.06 mm (it reads 0.06 when closed). Corrected reading == observed −- zero error:

34.54−0.06=34.48 mm34.54 - 0.06 = 34.48\ \text{mm}

Answer: LC = 0.02 mm; diameter = 34.48 mm.

(c) Micrometer

Least count =pitchthimble divisions=0.550=0.01= \dfrac{\text{pitch}}{\text{thimble divisions}} = \dfrac{0.5}{50} = 0.01 mm.

Observed reading =12.5+27×0.01=12.77= 12.5 + 27 \times 0.01 = 12.77 mm.

The zero error is −0.02-0.02 mm, so the correction is +0.02+0.02 mm:

12.77−(−0.02)=12.79 mm12.77 - (-0.02) = 12.79\ \text{mm}

Answer: corrected micrometer reading = 12.79 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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