ASVAB Shop Information Practice Test 4

Answer: C

Correct answer: A blind hole 5 mm in diameter and 12 mm deep. Read each part of the callout: Ø5 is a 5 mm diameter, and DEPTH 12 makes it a blind hole that ends 12 mm into the part. A through-hole would be marked THRU.

• A: The Ø symbol is in front of the 5, so the diameter is 5 mm, and DEPTH means the hole does not go through.
• B: A quantity would be shown with a mark such as 12X, not with the word DEPTH.
• C (correct): Ø gives the 5 mm diameter, and DEPTH 12 means the hole stops 12 mm into the material.
• D: Ø means diameter, not radius; a radius would be marked with R.

Answer: B

Correct answer: Center punch. A center punch is struck lightly at the intersection of the layout lines to leave a small indentation that guides the drill point. Check the layout dimensions before punching, because a misplaced punch mark is hard to correct.

• A: A try square lays out or checks a 90-degree line; it does not make a starting dent for a drill.
• B (correct): It leaves a small cone-shaped dent that gives the drill point a starting seat at the marked location.
• C: Feeler blades check the size of a narrow gap; they cannot mark a hole location.
• D: A dial indicator measures small changes in position such as runout; it does not mark the work.

Answer: C

Correct answer: Shearing. In shearing, two blades pass each other closely, like scissors, applying enough force to fracture sheet metal along a line without producing chips. Hand snips, squaring shears, and power shears all use this principle. The key clue is opposing blades on flat sheet. Turning and reaming are chip-producing machining operations on round work, and brazing joins metal with a filler rather than separating it.

• A: Brazing is a joining process with a filler metal.
• B: Turning removes chips from rotating workpieces, often cylindrical stock.
• C (correct): Sheet shears separate material along the intended cut line by blade action.
• D: A reamer finishes the inside of an existing round hole rather than cutting a flat sheet apart.

Answer: B

Correct answer: An eight-hour time-weighted average requiring a hearing conservation program. OSHA averages a worker's noise exposure over a full shift rather than judging single loud moments. When that eight-hour average reaches 85 dBA, the employer must run a hearing conservation program, including exposure monitoring, annual audiograms, training and available hearing protectors. The permissible exposure limit is higher, 90 dBA, and the allowed time halves for every 5 dB increase. The trap is reading 85 as a per-tool rating or an instant ceiling.

• A: It is an exposure criterion, not a universal tool-performance rating.
• B (correct): The action level concerns a measured time-weighted worker exposure.
• C: Impulse or impact noise is addressed separately, with a 140 dB peak guideline, not by the 85 dBA action level.
• D: Instantaneous and duration-dependent effects cannot be simplified this way.

Answer: C

Correct answer: 0.004, or 0.4%. Engineering strain is the change in length divided by the original length: ΔL/L₀ = 0.4 mm ÷ 100 mm = 0.004. Because both lengths carry the same unit, strain is dimensionless, and multiplying by 100 expresses it as a percent. The most common slip is reporting the elongation itself, 0.4, as the strain and forgetting to divide. Strains in ordinary tensile tests are usually small fractions of a percent up to a few percent, so a value like 40% should look suspicious.

• A: This fails to divide the 0.4 mm elongation by initial length.
• B: This overstates the relative extension by a factor of one thousand.
• C (correct): Length change is 0.4 mm; dividing by original 100 mm gives 0.004.
• D: Final length itself is not strain; strain is fractional length change.

Answer: B

Correct answer: Curvature across the width of the board. Lumber warps in several named ways. Cup is a curve across the width, so the face looks like a shallow trough. Bow is a curve along the length of the face, crook is a curve along the edge, and twist means the corners no longer lie in one plane. Cupping often comes from uneven drying, since flatsawn boards shrink more on the side facing away from the center of the tree.

• A: Edgewise length curvature is usually described as crook.
• B (correct): Cupping curves the board across its width, lifting edges or center.
• C: Knots can occur on a flat board without causing cup by definition.
• D: Corners in different planes describe twist, not simply cup.

Answer: C

Correct answer: Crimping. A crimping tool squeezes the barrel of a terminal or connector so the metal deforms permanently around the wire strands, creating a tight mechanical and electrical connection without heat. It is the standard way to attach lugs and connectors in vehicle and electrical work. Brazing and stick welding both require melting metal, and adhesives rely on a material that cures, so none of them involves simply squeezing a sleeve onto a wire.

• A: Brazing uses molten filler and capillary wetting.
• B: Stick welding forms a weld with a coated electrode.
• C (correct): Crimping plastically deforms a compatible terminal to grip the conductor.
• D: Adhesive bonds depend on a curing or solidifying material.

Answer: D

Correct answer: 6 turns. Each full turn of a single-start thread advances the screw by one pitch, so the number of turns needed is distance divided by lead:

Check it by multiplying back, since 6 turns × 1.5 mm = 9 mm. A common error is multiplying distance by pitch, which gives 13.5 and mixes up the units. Another is confusing the travel in millimeters with the number of turns.

• A: Nine turns at a lead of 1.5 mm per revolution would advance 13.5 mm, exceeding the specified travel.
• B: Multiplying distance by pitch gives a number with incorrect unit logic.
• C: Three turns multiplied by the 1.5 mm-per-turn lead would advance only 4.5 mm, not the stated 9 mm.
• D (correct): Turns equal distance divided by lead: 9 mm ÷ 1.5 mm per turn = 6.

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: C

Correct answer: 1.0 Ω. Ohm's law gives the total resistance:

Subtract the bulb:

of extra resistance, such as from a corroded connection.

• A: This is the total resistance of the whole circuit, including the bulb, not the extra resistance alone.
• B: This is the bulb's own resistance; the question asks for what the wiring adds beyond it.
• C (correct): The circuit's total resistance is 12.0 / 3.0 = 4.0 Ω, and 4.0 minus the 3.0 Ω bulb leaves 1.0 Ω of extra resistance.
• D: This divides current by voltage, which flips Ohm's law; resistance equals voltage divided by current.