18 Precision Measurement & Part Inspection Practice Questions & Answers

1. 1. You are writing an inspection plan for a turned shaft. What is the first step?

   • A.Identify the critical dimensions and their tolerances on the print, since each tolerance decides which instruments have adequate discrimination✓ Answer
   • B.Lay out every micrometer, indicator and gage in the toolroom so the whole set is available at the bench
   • C.Machine the first ten parts and record the cycle time of each so the plan matches the production rate
   • D.Verify that the lathe spindle bearings and chuck run within the machine builder's runout specification

   The drawing drives the plan: until you know which features are critical and what tolerance each carries, you cannot choose instruments with adequate discrimination or write a repeatable sequence of checks. Tool selection, fixturing and the accept/reject criteria are all consequences of the print requirements, so they come after this step.

   Source: NIMS/ANSI 101-2001, Machining Level I, Duty 3.1 Part Inspection (develop an inspection plan from the part print)Report a problem with this question

2. 2. When choosing a measuring instrument to check a given dimension, the single most important factor is:

   • A.The material of the part and the type of cutting tool that produced the surface
   • B.The tolerance specified for that dimension, which sets the discrimination the instrument must have✓ Answer
   • C.How quickly the operator can take a reading during a production run
   • D.The overall length of the part and the way it will be clamped in the fixture

   The instrument must be able to resolve variation that is small compared with the tolerance being judged, so the tolerance sets the required discrimination. The gage maker's rule of ten asks for roughly one-tenth of the tolerance, and measurement standards accept a ratio no worse than about 4:1; material, speed and part size never override that requirement.

   Source: NIMS Machining Level I KSAO 6.2 (Precision Measuring Instruments); gage maker's rule of ten, ASME B89.7.3.1 test-uncertainty ratio guidanceReport a problem with this question

3. 3. A dimension is toleranced ±.0005 in. Applying the rule of ten, the minimum acceptable instrument is:

   • A.A dial caliper reading to .001 in., used with light contact pressure and read twice on every part
   • B.A steel rule graduated in 1/64 in., sighted with the eye directly over the graduation line
   • C.A 0–1 in. micrometer with a vernier sleeve reading to .0001 in., one-tenth of the .001 in. total tolerance✓ Answer
   • D.A 0–1 in. micrometer reading to .001 in. with the operator estimating to the nearest half division

   The total tolerance is .0005 + .0005 = .001 in., and one-tenth of that is .0001 in., so the instrument must discriminate to .0001 in. — a vernier micrometer. An instrument reading only .001 in. has the same discrimination as the entire tolerance band, and estimating half divisions is not a substitute for real graduations.

   Source: Gage maker's rule of ten (instrument discrimination ≈ 1/10 of the total tolerance); NIMS Machining Level I KSAO 6.2Report a problem with this question

4. 4. A print limits a journal diameter to .6240/.6255 in. The part measures .6238 in.; all other dimensions are in tolerance and the job ran at its target cycle time. The correct disposition is:

   • A.Accept it, because .0002 in. under the limit is within the normal measuring error of a micrometer
   • B.Accept it, because every remaining dimension conforms and the run met its production rate
   • C.Send it back to the lathe to be re-cut, since a diameter this close to the limit can still be corrected
   • D.Reject it — the diameter is below the low limit, and an undersize turned diameter cannot be salvaged by rework✓ Answer

   A part is a reject the moment any single dimension falls outside the limits on the print; conformance of other features, appearance and production rate have no bearing on the decision. An undersize turned diameter also cannot be salvaged by machining, because rework only removes more material.

   Source: NIMS/ANSI 101-2001, Machining Level I, Duty 3.1 (accept/reject decision against print tolerances; decimal accuracy ±.001 in.)Report a problem with this question

5. 5. Which method is used to measure the pitch diameter of a 60° external thread?

   • A.A thread micrometer with cone and V-anvils, or the three-wire method✓ Answer
   • B.An outside micrometer with flat anvils closed across the crests of the thread
   • C.A vernier caliper with the jaws set over the major diameter of the thread
   • D.A depth micrometer with its rod stepped down into the root of one thread groove

   Pitch diameter is the theoretical diameter at which the width of the thread and the width of the groove are equal, so it must be measured on the flanks — the job of a thread micrometer's cone and V contacts or of measuring wires seated in the grooves. Flat anvils or caliper jaws laid across the tops of the threads read the major (crest) diameter instead.

   Source: NIMS Machining Level I KSAO 6.2; ASME B1.1 (Unified inch screw threads — pitch diameter definition and measurement)Report a problem with this question

6. 6. What is the correct way to use a telescoping gage to check a bore?

   • A.Read the bore size directly from the graduated sleeve on the gage handle once the plungers have seated
   • B.Lock the plungers in the bore, withdraw the gage, then measure across the contacts with an outside micrometer✓ Answer
   • C.Preset the gage to the print size, force it into the bore and judge the size by how tightly it drags
   • D.Insert the gage and slip a feeler gage between one contact and the bore wall to find the difference

   A telescoping gage is a transfer instrument: it has no scale of its own, so it only captures the bore size mechanically and the value comes from the micrometer that measures the locked contacts. Rock the gage lightly through the bore centerline before locking so the contacts capture the true diameter rather than a chord.

   Source: NIMS Machining Level I KSAO 6.2 (transfer instruments: telescoping gages, small-hole gages, adjustable parallels)Report a problem with this question

7. 7. Before a dial bore gage can report the actual size of a bore, the gage must be:

   • A.Pressed firmly into the bore so the contacts compress and take any looseness out of the linkage
   • B.Fitted with the longest available anvil so the plunger stays at the end of its travel while reading
   • C.Zeroed on a master of the nominal bore size, such as a setting ring or a gage-block stack✓ Answer
   • D.Checked against the outside diameter of the same part with a micrometer so the readings can be averaged

   A dial bore gage is a comparator: its dial shows deviation from whatever master it was set on, not an absolute size, so bore size equals the master size plus or minus the indicated deviation. Once set, the gage is rocked through the bore and the lowest reading is taken, because that is the point where the contacts lie on a true diameter.

   Source: NIMS Machining Level I KSAO 6.2 (comparison instruments must be set to a known master before use)Report a problem with this question

8. 8. An inch micrometer with a vernier sleeve shows the sleeve exposed to the number 7 plus three more .025 graduations, the thimble on 12, and vernier line 4 aligned with a thimble line. The reading is:

   • A..7624 in.
   • B..7374 in.
   • C..7870 in.
   • D..7874 in.✓ Answer

   The spindle screw is 40 threads per inch, so one turn is .025 in.: the numbered sleeve division 7 is 7 × .100 = .700, the three extra graduations add 3 × .025 = .075, the thimble adds 12 × .001 = .012, and the aligned vernier line adds 4 × .0001 = .0004, giving .7874 in. Dropping the vernier gives .7870 and miscounting the .025 marks gives .7624 or .7374.

   Source: NIMS Machining Level I KSAO 6.1/6.2 — micrometer graduation (40 TPI spindle = .025 in./rev, thimble .001 in., vernier .0001 in.)Report a problem with this question

9. 9. A dial test (lever) indicator is set with its contact arm 30° away from parallel with the direction of travel. The indicator sweeps .010 in. With cos 30° = .866, the actual displacement of the surface is:

   • A..00866 in.✓ Answer
   • B..01155 in.
   • C..010 in. — the angle of the lever has no effect on the reading
   • D..005 in.

   A lever indicator measures along its contact arm, so when the arm is tilted the true movement is the indicated reading multiplied by the cosine of the angle: .010 × .866 = .00866 in. This cosine error always makes the indicator read high, which is why the contact should be kept as nearly parallel to the surface of travel as the setup allows.

   Source: NIMS Machining Level I KSAO 6.2 — dial test indicator cosine error (true value = indicated reading × cos θ)Report a problem with this question

10. 10. A hole is dimensioned 3/32 in. ±1/64 in. What should the GO and NO-GO members of the plug gage measure?

    • A.GO .1094 in., NO-GO .0781 in., since the GO member must pass the largest hole
    • B.GO .0781 in., NO-GO .1094 in. — the smallest and largest permissible hole✓ Answer
    • C.GO .0938 in. (the nominal size), NO-GO .1094 in.
    • D.GO .0781 in., NO-GO .0938 in. (the nominal size)

    Convert the fractions first: 3/32 = .09375 and 1/64 = .015625, so the hole limits are .078125 and .109375 in. The GO member is made to the smallest permissible hole (maximum material condition) so it proves every hole is at least that big, and the NO-GO member is made to the largest permissible hole so it must not enter.

    Source: Limit (GO/NO-GO) gaging convention and Taylor's principle; ASME B89.1.6 / ASME Y14.5 maximum material conditionReport a problem with this question

11. 11. Using a standard 81-piece inch gage block set, which stack builds 2.4387 in.?

    • A..1009 + .138 + .200 + 2.000
    • B..1007 + .148 + .190 + 2.000
    • C..1007 + .138 + .200 + 2.000✓ Answer
    • D..1007 + .238 + .100 + 2.000

    Build the stack right to left, eliminating the last digit first: .1007 clears the 7 and leaves 2.3380, .138 clears the .038 and leaves 2.2000, .200 leaves 2.000, and the 2 in. block finishes it. The other stacks either miss the target or call for a block the set does not contain — the 81-piece set holds .1001–.1009, .101–.149, .050–.950 in .050 steps and the 1, 2, 3 and 4 in. blocks, so .190 and .238 do not exist in it.

    Source: Standard 81-piece inch gage block set contents; ASME B89.1.9 gage block practice (right-to-left digit elimination, fewest blocks)Report a problem with this question

12. 12. Gage blocks are assembled into a stack by:

    • A.Lapping the mating faces with a fine abrasive compound just before each stack is assembled
    • B.Clamping them in a holder, which is what actually holds any stack of more than two blocks together
    • C.Magnetizing the blocks so they attract one another and stay aligned while the stack is handled
    • D.Wringing the cleaned faces together to displace the air film between them✓ Answer

    Wringing is a sliding-and-rotating motion that squeezes out the air between two lapped faces so molecular attraction and atmospheric pressure hold the blocks together as one length. Because every wrung joint adds a small error, a stack should use the fewest blocks that will make the size, with wear blocks placed on the ends.

    Source: ASME B89.1.9 (gage blocks — wringing practice); NIMS Machining Level I KSAO 6.3 Surface Plate InstrumentsReport a problem with this question

13. 13. A 10 in. sine plate must be set to 12°. Using sin 12° = .20791, the required gage block stack height is:

    • A.2.0791 in.✓ Answer
    • B.1.0396 in.
    • C.9.7815 in.
    • D.2.1256 in.

    The sine bar or plate forms a right triangle whose hypotenuse is the center distance between the rolls, so the stack height is H = L × sin θ = 10 × .20791 = 2.0791 in. Using the 5 in. length by mistake halves the answer to 1.0396, using cosine gives 9.7815 and using tangent gives 2.1256.

    Source: NIMS Machining Level I KSAO 6.3 / applied right-angle trigonometry — sine bar setting height H = L × sin θReport a problem with this question

14. 14. A steel part is measured with a micrometer immediately after machining, while still warm from the cut. The most likely result is:

    • A.It reads undersize, because the film of coolant on the surface draws the micrometer anvils closer together
    • B.It reads oversize, because the heated part has expanded and will shrink as it normalizes to 68 °F✓ Answer
    • C.The reading is unaffected, since steel expands far too little to matter over any shop temperature range
    • D.The reading is unaffected as long as the micrometer was zeroed on the same machine a few minutes earlier

    All dimensional standards, gage blocks included, are defined at the reference temperature of 68 °F (20 °C), and steel expands as it is heated, so a warm part measures larger than it will when it stabilizes. Good practice is to let the part and the instrument normalize on the surface plate, handle blocks and mics as little as possible, and wipe away coolant and chips before reading.

    Source: ISO 1 / ASME B89.1.9 — standard reference temperature for dimensional metrology is 20 °C (68 °F)Report a problem with this question

15. 15. A machined surface carries a finish symbol with the number 32. This specifies:

    • A.A roughness average of 32 thousandths of an inch (.032 in.) measured across the lay
    • B.A minimum roughness of 32 microinches, so any smoother surface would be out of specification
    • C.A maximum roughness average of 32 microinches Ra, one microinch being .000001 in.✓ Answer
    • D.A waviness height of 32 millimetres measured over the full length of the surface

    Inch surface texture callouts are roughness average in microinches, where 1 microinch = .000001 in., and the single number shown is a maximum unless a range is given. A smaller number therefore means a finer surface: 32 is finer than 63 general machining but coarser than a 16 ground finish.

    Source: ASME B46.1 (Surface Texture) — roughness average Ra stated in microinches; value in the symbol is a maximum unless a range is shownReport a problem with this question

16. 16. In the shop, the practical way to verify the roughness callout on a finished surface is to:

    • A.Project the edge of the part on an optical comparator and overlay the magnified profile on a chart
    • B.Sweep the face with a dial indicator mounted on a height gage and record the total indicator movement
    • C.Take several micrometer readings across the face and treat the spread between them as the roughness
    • D.Compare it with a surface roughness comparison specimen, or read Ra with a profilometer✓ Answer

    Roughness is the fine, closely spaced irregularity of the surface itself, so it is judged by sight-and-fingernail comparison against a calibrated comparison specimen or measured quantitatively by a stylus profilometer. Optical comparators check profile and form, and height gages, indicators and micrometers check size and geometry — none of them measures surface texture.

    Source: ASME B46.1 (Surface Texture) — roughness comparison specimens and stylus instruments; NIMS Machining Level I KSAO 6.2Report a problem with this question

17. 17. A hole is dimensioned .7500/.7512 in. and the mating shaft .7476/.7484 in. The allowance of this fit is:

    • A..0016 in., the smallest hole minus the largest shaft✓ Answer
    • B..0036 in., the loosest condition the two parts can reach
    • C..0012 in., which is the tolerance permitted on the hole
    • D..0008 in., which is the tolerance permitted on the shaft

    Allowance is the intentional minimum clearance, found at the tightest condition of the assembly: smallest hole minus largest shaft, .7500 − .7484 = .0016 in. Because that value is positive the parts always assemble with clearance; the .0036 in. figure is the maximum clearance, and .0012 and .0008 are the tolerances on the individual features, not the fit.

    Source: ANSI/ASME B4.1 (Preferred Limits and Fits for Cylindrical Parts) — allowance = minimum clearance = smallest hole − largest shaftReport a problem with this question

18. 18. A dimension is given as 2.375 ±.005 in. Converted to metric, it is:

    • A.60.325 ±0.005 mm
    • B.60.325 ±0.127 mm✓ Answer
    • C.60.325 ±0.0127 mm
    • D.93.50 ±0.197 mm

    One inch is exactly 25.4 mm, so the dimension becomes 2.375 × 25.4 = 60.325 mm and the tolerance must be converted with it: .005 × 25.4 = 0.127 mm. Leaving the tolerance in inches, or dividing by 25.4 instead of multiplying, are the two errors the wrong answers represent.

    Source: NIMS Machining Level I KSAO 6.4 (Metric Conversion); 1 in = 25.4 mm exactly (international inch definition)Report a problem with this question

Practice questions based on NIMS/ANSI 101-2001, Duties and Standards for Machining Skills Level I, and the published content of the NIMS Machining Level I theory exams, together with standard precision-machining practice. NIMS is a mark of the National Institute for Metalworking Skills; this site is not affiliated with or endorsed by NIMS. Machining Level I is a set of separate credentials, and most of them also require a hands-on performance test that this bank does not cover. The NIMS theory exams are open-reference, but questions here never depend on recalling a handbook table value, a citation number or a machine rating — always work from the print in front of you, your employer's written procedures, and the machine's own documentation, and confirm current requirements before testing. About the NIMS machining credentials →