Free Red Seal Industrial Mechanic (Millwright) (433A) Practice Test

Answer: A. Try to start it at the local controls, then return the controls to off

A lock only proves that a device was moved to off; it does not prove the right device was isolated. Attempting a start (the try-out or verification step), then returning the controls to off, confirms zero energy. Stored energy such as gravity, springs and belt tension must also be relieved or blocked.

Check it: RSOS task A-1

Answer: B. Parts can fracture and fly off under the high force a puller develops

A puller stores a large amount of energy in the screw, jaws and part. If a jaw slips or a hub or casting cracks, pieces can be thrown with great force. A blanket or guard and proper PPE reduce the risk, along with keeping the puller square and the jaws fully engaged.

Check it: RSOS task A-2

Answer: C. 10.2 mm

For metric threads the tap drill is about the major diameter minus the pitch: 12 − 1.75 = 10.25 mm, so the standard drill is 10.2 mm. A 12.0 mm drill is a clearance size and 8.5 mm is the tap drill for an M10 × 1.5 thread.

Check it: RSOS task A-3

Answer: C. 12.73 mm

Add the sleeve and thimble readings: 12.00 mm + 0.50 mm (the half-millimetre line) + 0.23 mm from the thimble = 12.73 mm. Forgetting the exposed half-millimetre line gives 12.23 mm, the most common reading error.

Check it: RSOS task A-5

Answer: A. So the acetone that stabilizes the acetylene is not drawn out through the valve

Acetylene is dissolved in acetone held in a porous filler inside the cylinder. If the cylinder lies down, liquid acetone can be withdrawn with the gas, damaging regulators and producing an unstable flame. A cylinder that has been on its side should stand upright for a period before use.

Check it: RSOS task A-6

Answer: C. About 1,155 kg

Leg load = total load ÷ (number of legs × sine of the angle from horizontal) = 2,000 ÷ (2 × 0.866) ≈ 1,155 kg. 1,000 kg would be true only for vertical legs; 1,414 kg is the result at 45°. Lower sling angles always increase leg tension.

Check it: RSOS task B-7

Answer: A. About 565 kg

Volume = 2.4 × 1.2 × 0.025 = 0.072 m³. Mass = 0.072 × 7,850 ≈ 565 kg. A common error is to use 10 mm instead of 25 mm (about 226 kg) or to double-count both faces. Always confirm the load weight before selecting rigging.

Check it: RSOS task B-7

Answer: C. U-bolt on the dead (short) end and the saddle on the live (load) end

The saddle supports the rope without crushing it, so it goes on the live end that carries the load; the U-bolt bears on the dead end. The trade saying is never saddle a dead horse. Follow the clip maker's number of clips, spacing and torque, and retighten after the first load.

Check it: RSOS task B-8

Answer: A. Any worker who sees a hazard, and the operator must obey it

Normal signals come from one designated signaller so the operator is not confused. The emergency stop is the exception: anyone who sees danger may give it, and the operator must stop immediately. The lift resumes only after the hazard is cleared and the designated signaller resumes control.

Check it: RSOS task B-8

Answer: B. 1,200 rpm

Synchronous speed = 120 × frequency ÷ number of poles = 120 × 60 ÷ 6 = 1,200 rpm. An induction motor runs slightly below this because of slip, so a six-pole motor's nameplate typically shows a speed a little under 1,200 rpm.

Check it: RSOS task C-9

Answer: A. 40 mm

For standard metric bearings with a bore code of 04 or higher, the last two digits multiplied by 5 give the bore in millimetres: 08 × 5 = 40 mm. Codes 00, 01, 02 and 03 are the exceptions at 10, 12, 15 and 17 mm.

Check it: RSOS task C-10

Answer: D. Intermediate-shaft yokes in phase and equal operating angles at both joints

A single universal joint at an angle speeds up and slows down twice per revolution. A second joint cancels this only when the yokes on the intermediate shaft are in the same plane (in phase) and both joints run at equal angles. Out-of-phase yokes or unequal angles cause vibration and wear.

Check it: RSOS task C-11

Answer: A. 750 rpm

Driven speed = driver speed × driver diameter ÷ driven diameter = 1,750 × 150 ÷ 350 = 750 rpm. Inverting the ratio gives 4,083 rpm, which would mean the larger sheave turns faster, an obvious check that the answer is wrong.

Check it: RSOS task C-12

Answer: C. About 133 rpm

First stage: 1,200 × 20 ÷ 60 = 400 rpm. Second stage: 400 × 15 ÷ 45 ≈ 133 rpm. The overall ratio is 3 × 3 = 9 : 1 reduction. Stopping after the first stage gives 400 rpm; multiplying the ratios the wrong way gives 10,800 rpm.

Check it: RSOS task C-13

Answer: B. 0.20 mm

A rim indicator rotating around the hub sees the offset on both sides of the centreline, so its total indicator reading is twice the actual offset. 0.40 mm ÷ 2 = 0.20 mm. The sign of the reading and how the indicator is mounted then tell you which shaft is high.

Check it: RSOS task C-14

Answer: A. Remove 0.11 mm of shims

Position at the rear feet = −0.15 + 0.04 × 6.5 = +0.11 mm, so the rear feet are 0.11 mm high and shims must be removed. (At the front feet: −0.15 + 0.04 × 2.5 = −0.05 mm, so add 0.05 mm.) Adding the offset and slope without signs gives the wrong 0.41 mm.

Check it: RSOS task C-14

Answer: B. About 33%

Under the fan laws, flow varies with speed, pressure with speed squared and power with speed cubed: 1.1³ ≈ 1.33, an increase of about 33%. A modest speed change can overload the motor, so check the motor current after any sheave change. 21% is the pressure increase.

Check it: RSOS task D-16

Answer: B. The NPSH available has fallen below what the pump requires, so it is cavitating

A lower suction level reduces the static head on the suction side, lowering the net positive suction head available. When it drops below the pump's NPSH required, vapour bubbles form and collapse in the impeller, producing a gravel or marbles sound, unsteady pressure and pitting damage.

Check it: RSOS task D-17

Answer: D. About 33 m

Head varies with the square of speed: 50 × (1,450 ÷ 1,780)² ≈ 50 × 0.664 ≈ 33 m. 41 m is a straight-ratio answer (flow law), and 27 m is the cube law, which applies to power.

Check it: RSOS task D-17

Answer: A. About 7.9 : 1

Compression ratio uses absolute pressures: (700 + 101.3) ÷ 101.3 ≈ 7.9. Dividing gauge pressure by atmospheric gives 6.9, a common mistake. High ratios raise discharge temperature, which is why multistage machines use intercoolers.

Check it: RSOS task D-18

Answer: D. It moves toward the end of the idler it contacts first

The basic training rule is that a belt moves toward the end of an idler it contacts first. Training is done by shifting idler ends slightly, a little at a time, over several idlers ahead of the problem area, after checking that the structure is square and the load is centred.

Check it: RSOS task D-20

Answer: B. About 110 kN

Area = π × (0.05 m)² ≈ 0.007854 m². Force = pressure × area = 14,000,000 Pa × 0.007854 m² ≈ 110,000 N, or about 110 kN. Using the diameter instead of the radius gives four times the area, about 440 kN.

Check it: RSOS task E-21

Answer: D. About 133 mm/s

Speed = flow ÷ area. Flow = 40 L/min = 0.000667 m³/s; area = π × 0.04² ≈ 0.005027 m². Speed ≈ 0.133 m/s, or 133 mm/s. Retraction is faster for the same flow because the rod reduces the effective area.

Check it: RSOS task E-21

Answer: D. Dry nitrogen

Dry nitrogen is inert, so it will not support combustion or react with the oil. Oxygen or compressed air under high pressure with hydraulic oil can cause an explosion (dieseling). Precharge is checked with the fluid side drained and the system isolated.

Check it: RSOS task E-21

Answer: D. Filter, regulator, lubricator

Air is filtered first to protect the regulator from dirt and water, then regulated to the working pressure, and finally lubricated (where the components require it) so the oil mist travels downstream to the valves and cylinders.

Check it: RSOS task E-22

Answer: B. To vent the cylinder's air at the port so the piston moves faster

Without it, exhaust air must travel back through the line and directional valve. A quick exhaust valve dumps that air to atmosphere right at the cylinder port, increasing piston speed. Restricting exhaust air (meter-out) is what a flow control does to slow a cylinder.

Check it: RSOS task E-22

Answer: C. Rotor unbalance, such as uneven dirt build-up on the blades

Unbalance produces a strong radial 1× peak with relatively low axial vibration. Misalignment usually shows high axial vibration and 2× components; bearing defects appear at non-synchronous frequencies; gear problems appear at gear mesh frequency. Cleaning and balancing the rotor is the usual fix.

Check it: RSOS task F-23

Answer: D. 690 Hz

Gear mesh frequency = number of teeth × shaft speed in revolutions per second = 23 × (1,800 ÷ 60) = 23 × 30 = 690 Hz. 41,400 is the same product in cycles per minute, which must be divided by 60 to get hertz.

Check it: RSOS task F-23

Answer: B. Kinematic viscosity of about 68 mm²/s (cSt) at 40 °C

ISO viscosity grades classify industrial oils by their midpoint kinematic viscosity in mm²/s (centistokes) at 40 °C. Engine oils use the separate SAE system, and greases are graded for consistency by NLGI numbers, which run from 000 to 6.

Check it: RSOS task F-23

Answer: C. Direction of rotation, by bumping the motor briefly

Running a pump backward can loosen threaded impellers or damage seals, so rotation is checked with the motor uncoupled and the area clear. Hot alignment can only be checked after the machines run together at operating temperature.

Check it: RSOS task F-24