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Chapter 4 · 2 hours

Measuring and Gauging

Practice questions

Practice questions and answers

2 exam-style questions on this chapter, written for this site from the official syllabus. We haven’t found past IOE papers for this subject yet; if you have some, share them in the community.

  • Practice · 4+4 marks

(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.

Answer

(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.

  • Practice · 3+3 marks

What are slip gauges (gauge blocks)? Explain wringing. Using a set of 87 pieces (1.001 to 1.009 in steps of 0.001 mm; 1.01 to 1.49 in 0.01 mm; 0.5 to 9.5 in 0.5 mm; 10 to 100 in 10 mm), select the minimum number of blocks to build 48.325 mm.

Answer

Slip gauges

Slip gauges are rectangular blocks of hardened steel, carbide or ceramic, lapped flat and parallel to a very high accuracy, with the working faces a precise distance apart. They are the working standard of length in workshops, used to calibrate other gauges, set sine bars, comparators and for precision inspection. Grades: 00 (calibration), 0 (inspection), 1 (workshop).

Wringing

Wringing is sliding two clean blocks together with light pressure and a twisting motion so they stick together.

  • The faces are lapped so flat that a thin film of oil or moisture and molecular attraction hold them with a force of many newtons; the joint thickness is nearly zero (about 0.01 micron).
  • Procedure: clean the faces, place one block across the other at right angles, slide and rotate to align.
  • It lets several blocks be combined to give any size, with errors adding only slightly.

Building 48.325 mm (87-piece set)

Remove the decimal places from the last digit upwards, each time using one block.

StepRemaining sizeBlock taken
148.3251.005 (removes the 0.005)
247.3201.32 (removes 0.32)
346.0006.0
440.00040.0

Check: 1.005 + 1.32 + 6.0 + 40.0 = 48.325 mm.

Answer: four blocks - 1.005 + 1.32 + 6.0 + 40.0 = 48.325 mm.

Written from the official syllabus. Questions and answers are written for this site; check them against your class notes.

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