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Chapter 4 · 3 hours

Linear Measurement

Practice questions

Practice questions and answers

3 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 · 3+3+2 marks

(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.

Answer

(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.

  • Practice · 8 marks

What are slip gauges? Explain wringing, the grades of slip gauges and the precautions in their use. Build up a size of 58.275 mm from a set containing 1.005 mm; 1.001 to 1.009 mm in steps of 0.001 mm; 1.01 to 1.49 mm in steps of 0.01 mm; 0.5 to 9.5 mm in steps of 0.5 mm; and 10 to 100 mm in steps of 10 mm, using the minimum number of gauges.

Answer

Slip gauges (gauge blocks) are rectangular blocks of hardened, stabilised steel (or ceramic, tungsten carbide) with two opposite faces flat and parallel to a very high accuracy and a precise distance apart. They are the working end standards of length used to calibrate instruments, set comparators and sine bars, and for direct precision measurement.

Wringing

When two clean gauge faces are slid together with slight pressure and a twisting motion, they stick together. The adhesion is due to a thin film of oil or moisture (molecular attraction and atmospheric pressure). Wringing removes the air gap, so the length of the stack is the sum of the individual sizes with no measurable error. Gauges must be wrung just before use and separated soon after.

Grades (IS 2984 / BS 4311)

  • Grade 00 (reference/calibration) for calibration laboratories.
  • Grade 0 for inspection and master use.
  • Grade 1 for workshop inspection and setting tools.
  • Grade 2 for general workshop use and tool room.

Precautions

Keep at standard 20∘20^\circC and let the gauges soak; hold them with minimum contact so that body heat is not conducted; clean with a soft lint-free cloth and a solvent; wring correctly; coat with petroleum jelly or rust inhibitor after use; avoid dropping; use the minimum number of gauges to reduce cumulative error; use protector gauges at the end of stacks that are used for measuring.

Building 58.275 mm

Always eliminate the last decimal place first.

StepGauge chosenRemainder
Start58.275
11.00557.270
21.2756.000
36.050.000
4500

Check: 1.005+1.27+6.0+50=58.2751.005 + 1.27 + 6.0 + 50 = 58.275 mm.

Answer: 58.275 mm = 1.005 + 1.27 + 6.0 + 50 mm (4 gauges).

  • Practice · 8 marks

Describe the construction and working of an external micrometer. Briefly describe the internal micrometer, depth micrometer, vernier height gauge and vernier depth gauge with one application each. List the common errors of a micrometer.

Answer

External micrometer

It works on the principle of a screw and nut. A precision screw (spindle) of pitch 0.5 mm turns inside a fixed nut in the barrel (sleeve). A thimble attached to the spindle is divided into 50 parts, so one division moves the spindle 0.5/50=0.010.5/50 = 0.01 mm.

   anvil  spindle   sleeve  thimble  ratchet
  |==|   |=========|-----|=|||||||||=[R]
  |  |___|         |
  |__________frame__________|

Parts: U-shaped frame, fixed anvil, spindle, locking nut, sleeve with datum line and the main scale (0.5 mm marks), thimble, and a ratchet stop that gives constant measuring force (about 5 to 10 N) to avoid error due to excessive pressure.

Reading: sleeve reading + thimble division on datum line ×\times 0.01 mm. Used for outside diameters and thicknesses to 0.01 mm (vernier micrometers read 0.001 mm).

Other instruments

InstrumentPrinciple / use
Internal micrometerSame screw principle, with jaws or rods (extension rods) that touch the bore wall; measures hole diameters, e.g. cylinder bores
Depth micrometerA base (bridge) rests on the work face and a rod projects from the spindle; measures depth of slots, holes, steps
Vernier height gaugeA vernier caliper mounted vertically on a heavy base with a scriber; marks and measures heights on a surface plate, used in layout work
Vernier depth gaugeA beam with a vernier scale and a base; measures depth of holes and recesses with least count 0.02 mm

Common errors of a micrometer

  1. Zero error (anvil and spindle in contact do not read zero).
  2. Wear of anvil and spindle faces and of the screw thread; backlash.
  3. Non-parallelism or flatness error of measuring faces.
  4. Error in the pitch of the screw and thimble graduation.
  5. Variable measuring force if the ratchet is not used; deflection of the frame.
  6. Temperature effects, e.g. holding the frame in the hand, and parallax in reading.

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

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