Shop Information Practice Test 2

Answer: B

Correct answer: Extension lines. Extension lines carry the edges of a feature out to the dimension line, so the number clearly applies between those two points. Leaders point notes at features, and hidden lines show unseen edges.

• A: A leader connects a note or callout to a feature; it does not mark the ends of a measured distance.
• B (correct): Extension lines project from the feature so the reader knows which points or surfaces the dimension spans.
• C: Hidden lines are short dashes showing edges that cannot be seen from that view.
• D: A cutting-plane line shows where an imaginary cut is made for a section view.

Answer: A

Correct answer: A hex key that fits inside the head. A socket head has an internal hexagon recess, so it is driven with a hex key (Allen key) or a hex bit inserted into the head.

• A (correct): The socket head has an internal hex recess that accepts a hex key or hex bit.
• B: The outside of a socket head is round, so a wrench has no flats to grip.
• C: A flat blade fits a slotted head, not an internal hex recess.
• D: Gripping the threads would damage them and is not how any cap screw is driven.

Answer: B

Correct answer: Dowel pin. A dowel pin is a precisely sized cylindrical pin fitted into matching holes in two parts so they line up the same way every time they are assembled. Bolts or screws usually supply the clamping force, while the dowels provide accurate location and resist sideways shifting. The other choices are consumables for welding, soldering or gluing and have no precise fit. A useful cue: when a question says align or locate, think pins.

• A: A welding electrode is not intended as a locating pin.
• B (correct): A dowel pin primarily locates components, although details depend on design.
• C: A mixing nozzle dispenses adhesive, not a metal locating dowel.
• D: Solder is not a precision locating pin.

Answer: C

Correct answer: A rounded peen opposite a flat striking face. A ball-peen hammer, also called a machinist's hammer, has a flat face for striking punches and chisels and a rounded ball on the other end. The ball is used for peening, which means shaping metal, setting rivets and rounding over edges. Its hardened steel head suits metalwork rather than carpentry. The most tempting confusion is the claw hammer, whose split claw pulls nails from wood; the name ball-peen itself points to the ball shape.

• A: Nail-pulling claws distinguish claw hammers.
• B: Saw blades are separate cutting tools.
• C (correct): A rounded peen identifies the typical metalworking pattern.
• D: Toothed jaws are characteristic of pipe wrenches.

Answer: C

Correct answer: 0.10 mm. Total indicated variation, also called total indicator runout, is the full sweep of the needle from its lowest to its highest reading during one rotation. Because the readings sit on both sides of zero, the range includes both distances: 0.03 mm below zero plus 0.07 mm above zero equals 0.10 mm. Subtracting the sizes of the two numbers, giving 0.04 mm, ignores that the needle passed through zero.

• A: This triple-counts the excursion and has no basis in the endpoint readings.
• B: Subtracting magnitudes loses the negative-to-positive excursion.
• C (correct): The peak-to-peak swing equals maximum minus minimum: 0.07 - (-0.03) = 0.10 mm.
• D: This uses only the upper endpoint rather than the full range.

Answer: A

Correct answer: A router with a profile bit. The key clue is a shaped edge on work that stays still. A router's rotating bit forms the profile, whereas a lathe rotates the work and a jointer or planer only produces flat surfaces.

• A (correct): A router spins a shaped bit along the edge of stationary work to form decorative profiles, grooves, and rabbets.
• B: A lathe spins the workpiece itself to make round parts like legs or bowls; it cannot spin a whole tabletop to profile its edge.
• C: A jointer flattens a face or straightens an edge, even with its fence set; it does not form a shaped decorative profile.
• D: A thickness planer makes a board an even thickness; it does not shape edges.

Answer: C

Correct answer: Stop use and have it repaired through authorized procedures. Insulation is what keeps current inside the conductors; once it is broken, touching the plug while it is energized can send current through the body, and a damaged plug may also lose the ground connection that protects against faults. A tool that still runs proves only that the circuit is complete, not that it is safe. The tool should be unplugged, tagged out of service and repaired or replaced by a qualified person.

• A: Water may worsen electrical exposure and create slipping hazards.
• B: Motor and switch operation do not prove that exposed conductors are safe to touch.
• C (correct): Damaged insulation or exposed conductors can cause shock, so the tool must be taken out of use and properly repaired.
• D: A rag does not repair conductor insulation or plug integrity.

Answer: A

Correct answer: 8 turns. On a single-start thread, lead equals pitch, so each turn moves the screw 1.5 mm.

• A (correct): For a single-start thread the screw advances one pitch per turn, so 12 / 1.5 = 8 turns.
• B: This multiplies 12 by 1.5 instead of dividing the distance by the advance per turn.
• C: This divides the pitch by the distance, which inverts the ratio.
• D: This assumes 1 mm of travel per turn instead of the stated 1.5 mm pitch.

Answer: D

Correct answer: 0.9 mm. Step 1: stress = 24,000 N ÷ 200 mm² = 120 MPa.

Step 2: strain = stress ÷

MPa = 0.0006.

Step 3: stretch = 0.0006 × 1,500 mm = 0.9 mm. Keep E in MPa and the length in millimeters.

• A: This divides 120 MPa by 200 without converting GPa to MPa, giving an impossible strain of 0.6.
• B: The strain of 0.0006 is right, but it was multiplied by 1.5 (meters) and then labeled as millimeters.
• C: This uses 20,000 MPa for the modulus, a slip of one zero in the GPa-to-MPa conversion.
• D (correct): Stress is 120 MPa, strain is 120 ÷ 200,000 = 0.0006, and 0.0006 × 1,500 mm = 0.9 mm.

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.