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Vernier caliper and micrometer readings

Short answer

Vernier least count = 1 – 49/50 = 0.02 mm; correct reading = 34 + 27×0.02 − 0.06 = 34.48 mm. Micrometer least count = 0.5/50 = 0.01 mm; correct reading = 7.5 + 32×0.01 + 0.02 = 7.84 mm.

(a) A vernier caliper has 50 divisions on the vernier scale equal to 49 mm on the main scale (1 mm divisions). Find its least count. The main scale reads 34 mm to the left of the vernier zero, the 27th vernier line coincides with a main-scale line, and the instrument has a zero error of +0.06 mm. Find the correct reading. (b) An outside micrometer has a screw pitch of 0.5 mm and 50 divisions on the thimble. The sleeve shows 7.5 mm and the thimble reading line is the 32nd division. The zero error is -0.02 mm. Find the correct reading.

(a) Vernier caliper

Least count (LC) = 1 main-scale division - 1 vernier division.

1 VSD=4950=0.98 mmLC=1−0.98=0.02 mm\begin{aligned} 1\ \text{VSD} &= \frac{49}{50} = 0.98\ \text{mm} \\ \text{LC} &= 1 - 0.98 = 0.02\ \text{mm} \end{aligned}

Reading = main scale reading + (coinciding division x LC)

Observed=34+27×0.02=34.54 mmCorrect=Observed−zero error=34.54−0.06=34.48 mm\begin{aligned} \text{Observed} &= 34 + 27 \times 0.02 = 34.54\ \text{mm} \\ \text{Correct} &= \text{Observed} - \text{zero error} = 34.54 - 0.06 = 34.48\ \text{mm} \end{aligned}

A positive zero error means the instrument reads +0.06 mm when closed, so it is subtracted.

Answer (a): LC = 0.02 mm; correct reading = 34.48 mm.

(b) Micrometer

LC=pitchthimble divisions=0.550=0.01 mm\text{LC} = \frac{\text{pitch}}{\text{thimble divisions}} = \frac{0.5}{50} = 0.01\ \text{mm} Observed=7.5+32×0.01=7.82 mmCorrect=7.82−(−0.02)=7.84 mm\begin{aligned} \text{Observed} &= 7.5 + 32 \times 0.01 = 7.82\ \text{mm} \\ \text{Correct} &= 7.82 - (-0.02) = 7.84\ \text{mm} \end{aligned}

A negative zero error means the closed micrometer reads -0.02 mm, so the correction is added.

ItemValue
Sleeve reading7.50 mm
Thimble reading0.32 mm
Observed7.82 mm
Zero error-0.02 mm
Correct7.84 mm

Answer (b): LC = 0.01 mm; correct reading = 7.84 mm.

✓ Answer checked by Claude Opus 5.5 (high effort)

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