--- title: "Vernier caliper and micrometer readings" subject: "Metrology" chapter: 4 key: "semester_3/metrology:vernier-and-micrometer-readings" source: https://www.bctengineer.com/q/semester_3/metrology/vernier-and-micrometer-readings site: BCT Engineering --- # Vernier caliper and micrometer readings Metrology · Chapter 4: Linear Measurement - Asked in: year not given, 3+3+2 marks - Video: [How to Read a Metric Vernier Caliper](https://www.youtube.com/watch?v=vkPlzmalvN4) — WeldNotes - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_3/metrology/vernier-and-micrometer-readings - Whole chapter: https://www.bctengineer.com/md/semester_3/metrology/chapter-4 ## Question (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. ## 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. ## Answer #### (a) Principle and least count A vernier scale slides along the main scale. Its $n$ divisions equal $(n-1)$ 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: $$ \text{LC} = 1\ \text{MSD} - 1\ \text{VSD} = \frac{1\ \text{MSD}}{n} $$ The reading is: main scale reading (just left of the vernier zero) + (coinciding vernier division $\times$ LC). #### (b) Vernier caliper 1 VSD $= 49/50 = 0.98$ mm, so $$ \text{LC} = 1 - 0.98 = 0.02\ \text{mm} $$ Observed reading $= 34 + 27 \times 0.02 = 34.54$ mm. Zero error is positive: $+3 \times 0.02 = +0.06$ mm (it reads 0.06 when closed). Corrected reading $=$ observed $-$ zero error: $$ 34.54 - 0.06 = 34.48\ \text{mm} $$ **Answer: LC = 0.02 mm; diameter = 34.48 mm.** #### (c) Micrometer Least count $= \dfrac{\text{pitch}}{\text{thimble divisions}} = \dfrac{0.5}{50} = 0.01$ mm. Observed reading $= 12.5 + 27 \times 0.01 = 12.77$ mm. The zero error is $-0.02$ mm, so the correction is $+0.02$ mm: $$ 12.77 - (-0.02) = 12.79\ \text{mm} $$ **Answer: corrected micrometer reading = 12.79 mm.** ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. If n vernier divisions equal (n-1) main scale divisions, the least count is: A. n x 1 MSD B. 1 MSD / n C. 1 MSD x (n-1) D. 1 MSD / (n-1) Answer: B. LC = 1 MSD - 1 VSD = 1 MSD - (n-1)/n MSD = 1 MSD / n. (Key point: Vernier principle and least count) 2. With 1 mm main divisions and 50 vernier divisions coinciding with 49 mm, the least count is: A. 0.1 mm B. 0.05 mm C. 0.01 mm D. 0.02 mm Answer: D. 1 VSD = 49/50 = 0.98 mm, so LC = 1 - 0.98 = 0.02 mm. (Key point: Least count of caliper) 3. The closed jaws show the vernier zero 3 divisions ahead (+0.06 mm). The observed reading is 34.54 mm. The corrected reading is: A. 34.60 mm B. 34.54 mm C. 34.48 mm D. 34.51 mm Answer: C. Positive zero error is subtracted: 34.54 - 0.06 = 34.48 mm. (Key point: Zero error correction) 4. A micrometer (pitch 0.5 mm, 50 divisions) shows 12.5 mm on the sleeve and 27 on the thimble, zero error -0.02 mm. The corrected reading is: A. 12.77 mm B. 12.75 mm C. 12.79 mm D. 12.81 mm Answer: C. Observed 12.5 + 27 x 0.01 = 12.77 mm; subtracting the negative zero error (-0.02) adds 0.02, giving 12.79 mm. 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