Lesson Test: Drilling, Grinding and Shop Machines

Answer: A

Correct answer: Drill press. The key clue is that the tool spins while the workpiece stays still and clamped. A drill press holds the bit in a chuck on a vertical spindle and lowers it with a feed handle along the spindle axis. The common confusion is the lathe, which can also drill but normally spins the workpiece instead. A press brake and a surface grinder do not use twist drills at all; one bends sheet and the other abrades a surface flat.

• A (correct): A drill press drives a rotating drill and feeds it along the spindle axis toward restrained stock.
• B: A press brake bends sheet metal and does not use a rotating drill bit to form a hole.
• C: A surface grinder uses abrasive action on a supported workpiece, rather than primarily drilling holes.
• D: A conventional lathe usually spins the workpiece for turning, not the drill press operation described.

Answer: C

Correct answer: Engine lathe. A lathe is the machine that turns: the chuck spins the stock while a single-point tool on the carriage moves along or across it, removing metal to create diameters, steps and shoulders. Because the work rotates around one axis, the features it produces are round and concentric. The tempting mistake is the drill press, but there the tool rotates and the work stays still. Remember the pairing: lathe means rotating work, mill and drill mean rotating tool.

• A: A bench grinder uses abrasive wheels and is not the characteristic choice for turning a shaft shoulder.
• B: The typical drill press rotates the cutting bit and restrains the work.
• C (correct): A lathe rotates held stock and advances a tool to turn an outside diameter or shoulder.
• D: A power shear divides sheet along a line and does not turn a cylindrical shoulder.

Answer: C

Correct answer: Countersinking. A countersink tool has a cone-shaped cutting point, commonly 82 degrees for inch flat-head screws, and it cuts a matching cone at the top of the hole so a flat-head screw can sit flush. The word conical points to it. The closest wrong choice is counterboring, which makes a cylindrical step with a flat bottom for socket-head bolts. Reaming only refines the existing hole wall, and shearing does not cut holes at all.

• A: Counterboring creates a cylindrical larger-diameter recess instead of a conical one.
• B: Reaming refines an existing bore rather than intentionally forming a cone at its mouth.
• C (correct): Countersinking produces a conical mouth that can accommodate the head of a suitable fastener.
• D: Shearing cuts sheet along a line rather than forming a hole-mouth cone.

Answer: D

Correct answer: Milling with a suitable end mill. A milling machine rotates a multi-tooth cutter while the work, clamped to the table, is fed sideways into it. An end mill can cut on both its sides and its end, so traversing it along a path produces a slot with straight, parallel walls and a flat bottom. Drilling is the tempting choice, but a drill only cuts round holes along its own axis and cannot make straight side walls. Lathe facing and press-brake bending do not cut slots in plate.

• A: A press brake bends plate or sheet along a line; it does not remove material to form a slot.
• B: A drill cuts only a round hole along its axis and cannot produce straight slot walls by itself.
• C: Facing makes a flat surface on the end of rotating stock, not a slot in a plate.
• D (correct): A properly selected milling operation commonly cuts slots with straight sides in held plate.

Answer: C

Correct answer: Finish an existing hole to a close diameter. Drills leave holes that are slightly oversize, somewhat rough, and not perfectly round. A reamer is a multi-flute cutting tool that removes only a few hundredths of a millimeter (a few thousandths of an inch) from an existing hole to bring it to precise size and a smoother finish. That is why the usual sequence is to drill slightly undersize, then ream. The misconception is that a reamer can start a hole; it cannot, because it needs an existing bore to follow.

• A: A press brake performs bending; a reamer is an internal cutting tool.
• B: A threading die cuts external threads; a reamer works only inside an existing hole.
• C (correct): Reamers refine the size and finish of a hole that has already been drilled or bored.
• D: A reamer refines an internal hole; it does not drive nails by delivering an impact into wood.

Answer: A

Correct answer: Secure the work in a suitable fixture. When a drill breaks through or catches a burr, it can grab the work and spin it with the full torque of the spindle, turning a small part into a rotating blade. Clamping the work in a drill-press vise or fixture, or bolting it to the table, lets the machine absorb that force instead of your hand. Small parts are the most dangerous because they feel easy to hold. Also remove the chuck key before starting and keep guards in place.

• A (correct): Clamping prevents a grabbing drill from suddenly spinning the workpiece or ejecting it.
• B: Visibility does not justify bypassing guarding or other protective devices.
• C: Hands cannot reliably resist a bit grab and can be injured by spinning stock.
• D: A forgotten chuck key can be thrown when the spindle begins rotating.

Answer: B

Correct answer: A bonded wheel that removes metal with abrasive grains. A grinding wheel is made of countless hard abrasive grains, such as aluminum oxide or silicon carbide, held together by a bond. Each exposed grain acts like a tiny cutting edge, and as grains dull, they break away to expose fresh ones. That self-sharpening abrasive structure is what separates a grinding wheel from saw blades, threading tools and forming dies, which use defined teeth or shaped surfaces. Recognize the wheel by the words abrasive and bonded.

• A: That describes a band saw blade, not a grinder wheel.
• B (correct): Grinding removes stock with many abrasive grains acting as cutting points.
• C: Thread-rolling tooling forms threads by pressure and is not a bonded abrasive wheel.
• D: A forming die deforms sheet; it is not the grinder wheel described.

Answer: C

Correct answer: Remove it from service and have it replaced properly. An abrasive wheel spins at high surface speed and stores a large amount of energy, so a crack can let the wheel burst and throw fragments with great force. That is why wheels are inspected and ring-tested before mounting, and a damaged wheel is tagged out and replaced rather than repaired. Tape, adjusting the rest against a moving wheel, or speeding the wheel up all leave the hidden flaw in place and raise the chance of failure.

• A: Adjusting the rest against a moving wheel can damage the wheel and violates safe adjustment practice.
• B: Tape cannot restore the wheel’s designed structural strength, so grinding on it remains dangerous.
• C (correct): A cracked wheel presents a fragmentation hazard and must not be treated as usable.
• D: Higher speed adds stress to a cracked wheel and is not a safe inspection method.

Answer: A

Correct answer: Press brake. A press brake clamps sheet metal between a long punch and a matching die, then forces the punch down to fold the sheet along a straight line, making flanges, angles and channels. It is a forming machine, meaning it reshapes metal without removing chips. The distractors are all cutting tools: a drill press makes holes, a boring bar enlarges holes, and a band saw separates stock. If the task is bending sheet, think press brake.

• A (correct): A press brake is specifically associated with controlled sheet-metal bending.
• B: A drill press cuts axial holes into held work.
• C: A boring bar enlarges internal holes; it is not a sheet bending machine.
• D: A band saw separates stock with a toothed blade.

Answer: B

Correct answer: Band saw. A band saw blade is a long, flexible steel strip with teeth on one edge, welded into a loop that runs around two or more wheels. Because the blade moves continuously in one direction, it cuts smoothly and can follow curves or slice bar stock. The arbor press, surface grinder and lathe have no toothed blade: they press, abrade and turn. The name helps: a band is a continuous loop.

• A: An arbor press delivers controlled pressing force, not a continuously moving toothed blade.
• B (correct): A band saw carries an endless toothed blade around wheels to saw stock.
• C: A surface grinder uses abrasive grains on a wheel.
• D: A lathe uses a rotating workpiece and selected cutting tools.

Answer: A

Correct answer: 60 mm/min. Feed per revolution tells how far the drill advances each turn, and RPM tells how many turns happen each minute. Multiplying them gives distance per minute: 0.10 mm per revolution times 600 revolutions per minute equals 60 mm per minute. Checking units confirms it, because revolutions cancel and mm/min remains. The common errors are dividing the numbers or misplacing the decimal point, which produces answers off by a factor of ten or more.

• A (correct): Feed speed equals 0.10 mm/rev multiplied by 600 rev/min, giving 60 mm/min.
• B: This is ten times too small; the specified feed and RPM give 60, not 6.
• C: This exceeds the product of the given numbers by a factor of 100.
• D: Dividing feed per revolution by RPM reverses the requested relationship.

Answer: A

Correct answer: 30 seconds. Time equals distance divided by rate. Here 30 mm divided by 60 mm per minute gives 0.5 minute, and since one minute has 60 seconds, that is 30 seconds. A quick sanity check helps: the feed covers 60 mm in a full minute, so half that distance takes half a minute. Mistakes usually come from forgetting to convert minutes to seconds or from multiplying when you should divide.

• A (correct): Thirty millimeters divided by 60 mm/min is 0.5 minute, or 30 seconds.
• B: One minute at 60 mm/min covers 60 mm, twice the specified travel.
• C: Two minutes would cover 120 mm, four times the stated distance.
• D: Three seconds corresponds to only 3 mm at 60 mm/min.

Answer: B

Correct answer: 31.4 m/min. Cutting speed is the distance a point on the outer edge travels per minute. One revolution covers the circumference, π times the diameter: π × 10 mm ≈ 31.4 mm. At 1,000 RPM that is about 31,400 mm per minute, and dividing by 1,000 to convert to meters gives about 31.4 m/min. The usual errors are skipping the mm-to-m conversion or forgetting π, which shifts the answer by a factor of ten or about three.

• A: This is ten times the correct circumference-based linear speed.
• B (correct): π×10 mm×1000 rev/min divided by 1000 mm/m is approximately 31.4 m/min.
• C: RPM by itself is not linear distance without the circumference.
• D: This neglects a factor of ten in diameter or rotational conversion.

Answer: C

Correct answer: Lower RPM. Cutting speed depends on both diameter and RPM, because each revolution moves the outer edge one circumference, and circumference grows with diameter. A larger drill's edge travels farther per turn, so to keep the same surface speed you must turn it more slowly. This is why large drills run at low RPM and small drills at high RPM. The tempting mistake is thinking bigger tools need more speed; in fact excess speed overheats and dulls them.

• A: Higher RPM and a larger diameter both increase edge speed rather than keeping it constant.
• B: With greater diameter, the circumference and edge travel increase at unchanged RPM.
• C (correct): At constant peripheral speed, RPM is inversely proportional to tool diameter.
• D: Zero would eliminate rotation and cannot maintain the specified cutting speed.

Answer: A

Correct answer: 6.35 mm. The inch is defined as exactly 25.4 millimeters, so any inch measurement converts by multiplying by 25.4. One quarter inch is 0.25 inch, and

Because the definition is exact, the result is exact as well. A 6 mm drill is close but not the same size; it is a separate metric size. Another tempting error is 2.54 mm, which comes from confusing inches with centimeters and equals one tenth of an inch.

• A (correct): The inch is exactly 25.4 mm, so one quarter is 6.35 mm.
• B: Two and fifty-four hundredths millimeters equal one tenth of an inch, not one quarter of an inch.
• C: That is one whole inch rather than one quarter.
• D: A rounded metric drill nominal value is not the exact inch conversion.

Answer: B

Correct answer: A blind hole stops inside the material. The terms describe depth only. A through hole goes all the way through the part, so you can see light through it, while a blind hole is drilled to a set depth and has a bottom, often cone-shaped from the drill point. Size, threads and countersinks are separate features that either type of hole may or may not have. Blind holes need extra care when tapping because chips collect at the bottom and the tap can bottom out.

• A: Hole depth and hole diameter are independent specifications.
• B (correct): A blind hole has a bottom in the workpiece rather than passing completely through.
• C: A countersink is optional and is separate from blind-hole depth.
• D: Threads may be present but do not define whether a hole passes through.

Answer: D

Correct answer: Tapping. A tap is a hardened tool with threads ground on it; turning it into a drilled hole cuts or forms matching internal threads, such as those a bolt screws into. The hole is first drilled to a tap drill size slightly smaller than the bolt's major diameter, so the tap has material to cut. The matching tool for external threads on a rod is a die. Reaming is the tempting choice, but it only sizes and smooths a round hole.

• A: Reaming refines bore size or finish and does not typically make a helix.
• B: Shearing separates stock rather than forming a thread inside a hole.
• C: Facing cuts an end face rather than making internal screw threads.
• D (correct): A suitable tap creates or forms internal threads after correct hole preparation.

Answer: C

Correct answer: Milling. The clues are a rotating cutter and work fed sideways on a table, which is the definition of milling. Multi-tooth cutters such as end mills and face mills take repeated chips as the table moves, producing flats, slots and pockets. Turning is the main confusion: on a lathe the work spins and the tool is held still. Shearing and arbor pressing involve straight-line force and no rotating cutter.

• A: Shearing cuts with blades, not a rotary milling cutter.
• B: Turning usually involves a rotating workpiece rather than the described rotating cutter and table.
• C (correct): A milling machine commonly rotates a cutter and advances held stock or cutter to form slots and flats.
• D: An arbor press generates linear force for fitting rather than cutting a slot.

Answer: A

Correct answer: It moves work on a guided table to finish flat surfaces. A bench grinder is an offhand machine: you hold the work against the wheel for sharpening tools and removing burrs, so accuracy depends on your hand. A surface grinder clamps the work, often on a magnetic chuck, and moves it under the wheel on a table with precise down-feed, producing very flat, smooth surfaces to close tolerances. Both use abrasive wheels, which rules out saw blades, drills and bending. The key difference is controlled, guided motion.

• A (correct): Surface grinding uses controlled table or fixture motion to finish flat surfaces accurately.
• B: A surface grinder relies on abrasive wheel action, not a looped toothed saw blade.
• C: That is a drilling operation rather than the characteristic surface grinding task.
• D: Bending belongs to forming equipment such as press brakes.

Answer: B

Correct answer: 1/8 inch. The work rest supports the piece being ground. If the gap between the rest and the wheel is too wide, the work can be pulled down and wedged between them, which can jam the wheel, break it, or drag fingers in. OSHA therefore limits the work-rest opening on offhand grinders to 1/8 inch, and the gap must be readjusted as the wheel wears. The common confusion is the 1/4-inch limit, which applies to the upper tongue guard, not the rest.

• A: Half an inch is well above the specified work-rest maximum.
• B (correct): 29 CFR 1910.215(a)(4) sets a maximum one-eighth-inch work-rest opening for the covered offhand machines.
• C: One quarter inch relates to the distinct adjustable upper tongue opening in covered machines.
• D: Such an opening exceeds the applicable work-rest limit and increases jamming risk.

Answer: D

Correct answer: 1/4 inch. The tongue guard is the adjustable piece at the top of the wheel guard opening. Its job is to catch fragments thrown upward if the wheel breaks. OSHA allows no more than 1/4 inch between the tongue and the wheel periphery, and the tongue is moved in as the wheel wears smaller. Keep two numbers separate: 1/8 inch for the work rest below the work, 1/4 inch for the tongue guard above. Mixing up those two limits is the usual error.

• A: Three quarters of an inch exceeds the applicable tongue-guard clearance.
• B: One eighth inch is the distinct work-rest rule for covered offhand machines.
• C: A two-inch opening exceeds the OSHA maximum specified for the covered adjustable tongue.
• D (correct): 29 CFR 1910.215(b)(9) limits the specified upper tongue opening to one-fourth inch.

Answer: B

Correct answer: Only while the wheel is not moving. Adjusting the work rest means putting hands and tools close to the wheel and briefly loosening the rest. With the wheel turning, a slip could pinch a hand, wedge the rest into the wheel, or crack it. Stopping the wheel removes that risk, which is why the rule requires adjustment only with the wheel stationary. Spark visibility and operating speed have nothing to do with safe adjustment; the rule is simple: stopped wheel, then adjust.

• A: Combining work pressure and adjustment increases risk and violates the stopped-wheel principle.
• B (correct): The work-rest provision explicitly prohibits adjustment with the wheel in motion.
• C: Spark visibility is not a safe or permitted adjustment criterion.
• D: Adjustment during wheel motion is prohibited and risks contact.

Answer: B

Correct answer: It exceeds the maximum by 1/16 inch. To compare fractions, put them over a common denominator: 1/8 inch equals 2/16 inch, so a 3/16-inch gap is 1/16 inch larger than allowed and must be closed before grinding. In decimals, 0.1875 inch is greater than 0.125 inch. The tempting mistake is comparing the gap with the 1/4-inch figure, but that limit belongs to the upper tongue guard; each opening has its own maximum.

• A: The 1/4-inch limit concerns the upper guard opening, not the work rest.
• B (correct): 3/16 minus 2/16 equals 1/16 inch, so the work-rest opening is too wide.
• C: Three sixteenths equals 0.1875 inch, whereas one eighth is 0.125 inch, so these values cannot be equal.
• D: Three sixteenths is greater than two sixteenths, not smaller.

Answer: C

Correct answer: Produce a larger cylindrical recess around a hole. A counterbore cutter has a pilot that rides in the existing hole and a larger flat cutting end that machines a straight-walled, flat-bottomed step. That step lets a socket-head cap screw or bolt head sit at or below the surface. Watch the shape word: cylindrical means counterbore, conical means countersink. Threading and abrasive polishing are unrelated operations that do not form a recess around a hole.

• A: External threads are made by suitable dies or machine threading, not by counterboring.
• B: A conical recess is the defining geometry of a countersink, not a counterbore.
• C (correct): A counterbore leaves a flat-bottomed cylindrical relief around the hole opening.
• D: That is surface finishing, not counterboring.

Answer: D

Correct answer: Stop the machine and use a suitable chip-removal tool. Long, stringy chips are razor-sharp and can wrap around a turning spindle or cutter, so reaching in while the machine runs risks cuts and entanglement. Stopping the machine first eliminates the moving-part hazard, and using a brush, chip hook or pliers keeps hands away from sharp edges even when stopped. Compressed air is a poor substitute because it blows chips and fine particles toward eyes and other workers.

• A: Rotating cutters and sharp chips create entanglement and cut hazards.
• B: Higher spindle speed cannot replace deliberate and safe chip management.
• C: Airborne chips can strike workers and create additional exposure hazards.
• D (correct): Machine stoppage and a suitable brush or hook reduce contact with moving tools and sharp chips.

Answer: D

Correct answer: Force a suitably supported press-fit component into place. An arbor press is a hand-operated press that uses a lever and a gear rack to push a ram straight down with controlled force. It is used to press bearings, bushings, pins and gears onto shafts or into bores, or to push them out. It does not cut metal, which rules out grinding, boring and milling. Supporting the part correctly matters, because force through the wrong surface can crack or misalign it.

• A: That describes a suitable grinding operation.
• B: A boring tool or machine operation is not the same as arbor pressing.
• C: That is a milling operation, not a mechanical pressing operation.
• D (correct): An arbor press delivers controlled axial pressing force for appropriately fixtured jobs.

Answer: A

Correct answer: 12.06 mm. A tolerance of ±0.05 mm means the dimension may vary 0.05 mm above or below nominal. Subtracting and adding gives limits of 11.95 mm and 12.05 mm, and values exactly at those limits still pass. Checking each measurement against that range, only 12.06 mm falls outside, by 0.01 mm. The tempting error is rejecting 11.95 or 12.05 because they sit on the edge; limits are normally inclusive.

• A (correct): The accepted inclusive range is 11.95 to 12.05 mm; 12.06 exceeds the upper limit.
• B: 11.95 mm equals the lower limit and meets the stated symmetric tolerance.
• C: The nominal dimension sits at the center of the allowed interval.
• D: 12.05 mm equals the upper stated limit and is not outside it.

Answer: B

Correct answer: Departure of rotating geometry from its true axis. Runout measures how much a rotating surface wobbles relative to its axis of rotation. It is usually checked with a dial indicator touching the spindle, chuck or tool while it is turned slowly; the total swing of the needle is the runout. Excess runout makes a drill cut oversize, puts uneven load on cutter teeth and leaves a poor finish. Chip direction, hydraulic pressure and coolant volume have no connection to rotational accuracy.

• A: Chip fall direction does not define geometric rotational runout.
• B (correct): Runout describes eccentric or nonconcentric behavior during rotation and can affect machining.
• C: Hydraulic pressure is not a measurement of rotating eccentricity.
• D: Tank capacity is independent of spindle or tool concentricity.

Answer: D

Correct answer: Appropriate eye protection may still be needed beneath it. A face shield is secondary protection: it guards the face from larger particles and sparks, but it is open at the bottom and sides, so debris can still reach the eyes. Safety glasses or goggles worn under the shield are the primary eye protection. Personal protective equipment also does not replace machine guards or a properly rated, undamaged wheel. Think of it as layered protection, not one item doing every job.

• A: Wheel rating and integrity remain independent safety requirements.
• B: Shields reduce selected exposures but cannot prevent every particle pathway.
• C: Personal protection does not replace required guarding.
• D (correct): Flying particles can reach the eyes around a face shield; task-specific eye protection remains necessary.

Answer: B

Correct answer: Face milling. A face mill is a large-diameter cutter with several inserts around its face. It rotates with its axis perpendicular to the work surface, so the inserts sweep a wide path and leave a broad, flat surface in a few passes. Its name matches the feature: face milling produces faces. Drilling and tapping make holes and threads, and arbor pressing applies force without cutting.

• A: Drilling primarily produces holes, not broad flat external faces.
• B (correct): A suitably selected face mill is commonly used to produce broad flat surfaces.
• C: Tapping produces internal threads rather than a broad flat face.
• D: Pressing generates force but not the described chip-forming flat face.

Answer: A

Correct answer: 942 m/min. Rim speed is circumference times RPM. The wheel's circumference is π × 0.20 m ≈ 0.628 m, so at 1,500 RPM the rim travels about 0.628 × 1,500 ≈ 942 meters every minute. Because the diameter is already in meters, no unit conversion is needed. The usual traps are forgetting π, which gives 300, and treating RPM as a speed by itself. Rim speed matters because every wheel has a maximum safe speed.

• A (correct): π×0.20 m×1500/min is approximately 942 meters per minute of peripheral travel.
• B: This is about one tenth of the circumference-based result.
• C: RPM alone is not a linear speed and cannot be used directly as m/min.
• D: Diameter times RPM omits π from the circumference relationship.

Answer: C

Correct answer: 955 RPM. This reverses the cutting-speed formula. Cutting speed in m/min equals π × D × N ÷ 1,000 with D in millimeters, so

Here

and π × 10 ≈ 31.4, giving about 955 RPM. A quick estimate checks it: dividing by π is roughly dividing by 3, so 30,000 ÷ 30 ≈ 1,000. Forgetting π or the 1,000 conversion factor creates the wrong options.

• A: This ignores π and therefore overshoots the intended RPM by more than threefold.
• B: That confuses the linear speed input with rotational revolutions per minute.
• C (correct): Using N=1000v/(πD) gives 30,000/(π×10), about 955 RPM.
• D: A factor-of-ten reduction is inconsistent with the supplied diameter and speed.

Answer: D

Correct answer: 1,571 mm³. A through hole removes a cylinder, and cylinder volume is the area of the circle times the length: π × r² × h. Use the radius, which is half the 10 mm diameter, so r = 5 mm and the circle's area is π × 25 ≈ 78.5 mm². Multiplying by the 20 mm thickness gives about 1,571 mm³. The most common error is squaring the full diameter, which makes the answer four times too large.

• A: Diameter multiplied by depth omits the circular area πr².
• B: 314 mm³ is π × 10², which uses the diameter as the radius and ignores the 20 mm depth.
• C: This squares the diameter instead of the radius and gives four times too much.
• D (correct): The radius is 5 mm; π×5²×20 is approximately 1571 cubic millimeters.

Answer: C

Correct answer: 2 MPa. Pressure is force divided by the area it acts on. Here 1,200 N ÷ 600 mm² = 2 N/mm². Since 1 square meter contains 1,000,000 square millimeters, 1 N/mm² equals 1,000,000 N/m², which is 1 megapascal. So the answer is 2 MPa. The traps are multiplying instead of dividing, flipping the ratio, or assuming N/mm² equals kilopascals, which misses the unit conversion by a factor of 1,000.

• A: Multiplying the two given values rather than dividing is dimensionally wrong.
• B: Dividing area by force reverses the pressure ratio.
• C (correct): 1200/600 = 2 N/mm²; one newton per square millimeter equals one megapascal.
• D: 2 N/mm² equals 2 MPa, not 2 kPa; the units differ by a factor of 1000.

Answer: D

Correct answer: 9.525 mm. Convert the fraction to a decimal first:

Multiplying by exactly 25.4 mm per inch gives 9.525 mm. A useful shortcut is that 1/8 inch is 3.175 mm, so three eighths is three times that. The tempting error is 9.375, which just reuses the digits of the decimal inch value as if they were millimeters. Other options come from converting the wrong fraction, such as 1/8 or 5/8.

• A: 15.875 mm equals five eighths inch, not three eighths.
• B: 3.175 mm is one eighth inch, not three eighths.
• C: This incorrectly treats the decimal fraction as a metric numeric suffix.
• D (correct): 0.375 inches times exactly 25.4 mm/inch equals 9.525 millimeters.

Answer: D

Correct answer: Multiply mm/rev by rev/min to obtain mm/min. Feed per revolution, in mm/rev, describes how far the drill advances each turn; feed per minute, in mm/min, describes how fast it advances over time. Linking them requires the spindle speed in rev/min. Writing the units out shows why multiplication works: mm/rev × rev/min, the revolutions cancel, and mm/min remains. Adding values with different units, or dividing them in reverse, produces answers with meaningless units.

• A: They have different dimensions and cannot be added to produce a rate.
• B: This produces mismatched units and reverses the intended product.
• C: Multiplying by elapsed time does not yield a rotational frequency.
• D (correct): The revolution units cancel, leaving the axial feed rate in millimeters per minute.

Answer: A

Correct answer: 30 seconds. Divide distance by rate to get time: 45 mm ÷ 90 mm/min = 0.5 minute, which is 30 seconds. Because 45 is exactly half of 90, you can also reason that the drill covers half its one-minute distance in half a minute. The wrong options come from using the numbers directly as seconds or halving in the wrong direction. Always finish by converting minutes to seconds when the answer choices are in seconds.

• A (correct): 45 mm divided by 90 mm/min equals 0.5 min or 30 seconds.
• B: At 90 mm/min, 90 seconds covers 135 mm rather than 45 mm.
• C: At 90 mm/min, a 45-second interval would cover 67.5 mm, exceeding the requested 45 mm advance.
• D: At 90 mm/min, a 15-second interval covers only 22.5 mm, which is half the specified 45 mm.

Answer: C

Correct answer: It doubles. At a fixed RPM, the outer edge moves one circumference per revolution, and circumference equals π times diameter. Since circumference is directly proportional to diameter, doubling the diameter from 10 mm to 20 mm doubles the distance covered each turn and therefore doubles the edge speed. The tempting error is quadrupling, which confuses circumference with area; area scales with the square of diameter, but edge distance does not.

• A: At the same RPM a larger circle travels farther each revolution.
• B: A halving occurs if diameter halves at the same rotational speed.
• C (correct): Since edge speed is proportional to tool diameter at fixed RPM, doubling diameter doubles speed.
• D: The circumference scales linearly with diameter, not with its square.

Answer: B

Correct answer: The tongue opening exceeds its 1/4-inch maximum. Check each opening against its own limit. The work rest is set at 1/8 inch, which equals its 1/8-inch maximum and is acceptable. The upper tongue is at 3/8 inch, but its limit is 1/4 inch, which is 2/8 inch, so it is 1/8 inch too wide and must be moved closer. The typical errors are applying one limit to both openings or thinking a value equal to the maximum is a violation.

• A: The work rest and upper tongue are independently regulated geometrical clearances.
• B (correct): The work rest meets its 1/8-inch limit, but the 3/8-inch upper tongue exceeds the separate 1/4-inch maximum.
• C: Three eighths of an inch is larger than the upper tongue’s one quarter inch maximum.
• D: One eighth inch equals the maximum work rest clearance; it does not exceed it.

Answer: D

Correct answer: 7.97–8.03 mm. A bilateral tolerance of ±0.03 mm means subtract 0.03 mm from nominal for the lower limit and add 0.03 mm for the upper limit:

The total tolerance band is 0.06 mm, centered on nominal. Any measurement from 7.97 to 8.03 mm, inclusive, is acceptable. The wrong options shift the band to one side or misplace the decimal.

• A: This shifts the interval upward and fails to subtract tolerance from nominal.
• B: This uses a tenfold larger 0.30 mm tolerance instead of 0.03 mm.
• C: This shifts the interval downward instead of centering it around nominal.
• D (correct): Subtracting and adding 0.03 mm to 8.00 mm gives lower 7.97 and upper 8.03 mm.

Answer: B

Correct answer: 120 mm/min. Multiply feed per revolution by rotational speed so the revolution units cancel:

• A: This adds the RPM and the feed per revolution instead of multiplying them.
• B (correct): Feed per revolution times revolutions per minute gives 0.15 x 800 = 120 mm/min.
• C: This divides 800 by 0.15; the two quantities must be multiplied so revolutions cancel.
• D: This is a decimal-place slip; 0.15 x 800 is 120, not 12.

Answer: C

Correct answer: 15 seconds. First find the feed rate: 0.12 x 900 = 108 mm/min. Then time = travel / feed rate =

Convert to seconds: 0.25 x 60 = 15 seconds.

• A: This reads 0.25 minute as 25 seconds; a quarter of a minute is 15 seconds.
• B: This divides the feed rate by the travel (108 / 27) instead of travel by feed rate.
• C (correct): The feed rate is 108 mm/min, and 27 / 108 = 0.25 min, which is 15 seconds.
• D: This divides the travel by the RPM and skips the feed per revolution entirely.

Answer: B

Correct answer: 30.2 m/min. Edge speed equals pi times diameter times RPM:

/min.

Divide by 1,000 to convert to meters: about 30.2 m/min.

• A: This is the speed in millimeters per minute mislabeled as meters; divide by 1,000.
• B (correct): Circumference is about 50.27 mm; times 600 RPM is about 30,159 mm/min, or 30.2 m/min.
• C: This uses the 8 mm radius in place of the diameter in the pi x D formula.
• D: This multiplies diameter by RPM but leaves out pi, so it is not the circumference.

Answer: A

Correct answer: About 995 RPM. Rearrange the surface-speed relation:

With V = 25 m/min and D = 8 mm:

or about 995 RPM, before any real machine limits are considered.

• A (correct): RPM = 1000 x 25 / (pi x 8) = 25,000 / 25.13, which is about 995.
• B: This divides 25,000 by 8 and forgets pi, so it uses diameter instead of circumference.
• C: This multiplies pi x 8 x 25 instead of dividing the cutting speed by the circumference.
• D: This uses the 4 mm radius instead of the 8 mm diameter, doubling the answer.

Answer: B

Correct answer: Remove it from the chuck. The chuck key is used only with the machine stopped. Remove it before applying power, because spindle rotation can eject a key left in the chuck.

• A: The key only tightens the chuck; left in place it becomes a projectile once the spindle turns.
• B (correct): A key left in the chuck can be flung out when the spindle rotates, so it must be removed first.
• C: Tightening a chuck must never be done with the spindle turning.
• D: Loose tools on the machine can fall or be caught; the key should be removed and kept clear.

Answer: C

Correct answer: 20.03 mm. Calculate each limit separately: upper =

lower =

Only 20.03 mm lies within 19.99 to 20.04 mm.

• A: This would pass only if -0.04 were applied to the lower side; the lower limit is 19.99 mm.
• B: This is 0.02 mm below nominal, but only 0.01 mm undersize is allowed.
• C (correct): The limits are 19.99 mm to 20.04 mm, and 20.03 mm falls between them.
• D: This is 0.05 mm over nominal and exceeds the 20.04 mm upper limit.

Answer: A

Correct answer: About 1,696 mm³. Volume of a cylinder is $\pi r^{2}h$. The radius is half of 12 mm, so r = 6 mm.

• A (correct): With r = 6 mm, pi x 36 x 15 is about 1,696 mm³.
• B: This squares the 12 mm diameter instead of the 6 mm radius.
• C: This uses pi x diameter x thickness and never squares the radius.
• D: This squares the radius and multiplies by thickness but leaves out pi.

Answer: D

Correct answer: A die. A die is a hardened threaded ring with cutting edges on the inside. It is held in a diestock and turned onto the rod, cutting threads into the outside surface. A tap is the opposite: a threaded cutting tool turned into a drilled hole to make internal threads. A simple memory hook is that a nut needs a tap and a bolt needs a die. The end of the rod is usually chamfered first so the die starts square.

• A: A tap makes internal threads inside a prepared hole, not threads on the outside of a rod.
• B: A reamer refines the size and finish of an existing hole; it does not make threads.
• C: A countersink makes a conical recess at a hole opening.
• D (correct): A die cuts or forms external threads on rod or bolt stock.

Answer: B

Correct answer: 3 MPa. First find the contact area: 15 mm x 20 mm = 300 mm². Then pressure = force / area = 900 N / 300 mm² = 3 N/mm².

One N/mm² equals one MPa, so the average pressure is 3 MPa.

• A: This divides area by force (300 / 900); pressure is force divided by area.
• B (correct): The pad area is 15 x 20 = 300 mm², and 900 N / 300 mm² = 3 N/mm², which equals 3 MPa.
• C: This divides the force by one side length only (900 / 15) instead of by the area.
• D: This divides by the sum of the sides (15 + 20); area requires multiplying them.

Answer: C

Correct answer: Boring. A single tool cutting the inside diameter of a hole that already exists is boring. Facing works on the end surface, and turning works on the outside diameter.

• A: Facing cuts across the end of the work to make a flat end surface.
• B: Tapping makes internal screw threads; it does not simply enlarge the diameter.
• C (correct): Boring enlarges or refines an existing hole with a single-point tool on a bar.
• D: Knurling presses a grip pattern onto an outside surface rather than cutting an inside diameter.

Answer: D

Correct answer: Follow the machine's energy-isolation procedure. For servicing where a machine could energize unexpectedly, OSHA's hazardous-energy rule and the equipment instructions call for a defined isolation procedure. Simply switching off is not enough.

• A: An off switch alone may not prevent unexpected startup or stored-energy release during servicing.
• B: A person watching a switch is not a reliable substitute for proper isolation.
• C: Working on a belt while the machine runs exposes the worker to nip points and moving parts.
• D (correct): Covered servicing calls for isolating and controlling hazardous energy, not just turning a switch off.

Answer: D

Correct answer: A power shear. Match the job verb to the machine: cutting sheet along a line points to a power shear, while bending it into a flange would point to a press brake.

• A: A press brake bends sheet into angles and flanges; it does not cut the sheet apart.
• B: An arbor press pushes press-fit parts such as bushings into place.
• C: A surface grinder finishes flat surfaces with an abrasive wheel; it is not used to cut sheet to width.
• D (correct): A power shear cuts sheet along a straight line, which reduces its width.

Answer: A

Correct answer: Chatter. A boring bar acts like a diving board: the farther it extends from its holder, the more easily it bends and vibrates. When the cutting forces match that natural vibration, the tool bounces in and out of the cut, producing a high-pitched noise and evenly spaced marks on the surface. The usual fixes are reducing overhang, using a stiffer bar, or changing speed and feed. Runout is a different problem: an off-center rotation causes an uneven cut, not the noise and wavy pattern of chatter.

• A (correct): Chatter is self-excited vibration, often heard as noise and seen as a patterned finish; long overhang encourages it.
• B: Runout means rotation is off the intended axis; it is not mainly identified by squealing and a wave pattern.
• C: Burrs are raised edges left at a hole's entrance or exit, not a vibration pattern inside the bore.
• D: Glazing describes a dulled abrasive wheel surface, not vibration of a cutting tool.

Answer: A

Correct answer: 7.9375 mm. Convert the fraction to a decimal, then multiply by 25.4 mm per inch:

and 0.3125 x 25.4 = 7.9375 mm.

• A (correct): 5/16 inch is 0.3125 inch, and 0.3125 x 25.4 = 7.9375 mm exactly.
• B: This uses 25 mm per inch instead of the exact 25.4 mm.
• C: This multiplies by 2.54, which converts inches to centimeters, not millimeters.
• D: This divides 25.4 by 0.3125 instead of multiplying.