Auto Information Practice Test 6: Cooling and Diagnosis

Answer: B

Correct answer: It lowers the freezing point and raises the boiling point. Plain water freezes at 32°F, boils fairly easily, and lets metal rust. Adding glycol lowers the freezing point and raises the boiling point, and the inhibitors in the mix fight corrosion.

• A: Antifreeze is not added to thicken coolant; too much antifreeze actually reduces heat transfer.
• B (correct): Glycol widens the liquid range of the coolant, protecting against both freezing and boiling.
• C: Bearings are lubricated by engine oil; coolant never reaches them.
• D: Water is the better heat carrier; glycol is added for freeze and boil protection, not better heat transfer.

Answer: D

Correct answer: Hydroplaning, where water lifts the tires off the pavement. Hydroplaning occurs when the tread cannot push water away fast enough and a wedge of water lifts the tire off the road. Steering and braking grip nearly disappear. Higher speed, deeper water, worn tread and low pressure all raise the risk.

• A: Cupping is a wear pattern felt as noise or vibration, not a sudden loss of steering on water.
• B: Overloading causes flexing and heat, not a sudden lift-off on a wet road.
• C: Imbalance causes speed-related vibration, not loss of contact with the road.
• D (correct): When tires cannot clear water fast enough, a water wedge lifts them and grip nearly vanishes.

Answer: A

Correct answer: A short to power. When a wire touches another live circuit, it is a short to power: the device may run when switched off, or two circuits feed each other. A short to ground, by contrast, sends current straight to the body and usually blows the fuse. Naming the fault type guides the testing.

• A (correct): A wire touching another live circuit can feed a device so it runs when it should be off.
• B: A short to ground is a powered wire touching the body; that collapses resistance and usually blows the fuse instead of running the device.
• C: An open stops current, so the wiper would not run at all.
• D: Extra ground resistance would slow or weaken the wiper, not make it run with the switch off.

Answer: A

Correct answer: 22 ft/s². Use

The speed changes by 88 ft/s over 4 s.

• A (correct): Deceleration is the speed change divided by the time: 88 ft/s divided by 4 s is 22 ft/s squared.
• B: This multiplies speed by time, which gives a distance-like number, not a rate of slowing.
• C: This divides time by speed, which inverts the formula.
• D: This divides by 2 times the time, mixing in the 2 from the stopping-distance formula where it does not belong.

Answer: D

In a floating caliper, fluid pressure moves the single piston out against the inner pad. When that pad meets the rotor, the caliper body slides on its guide pins in the opposite direction and draws the outer pad onto the other face, so both pads clamp the rotor.

• A: The rotor is bolted to the hub and turns with the wheel; it does not shift sideways to meet a pad.
• B: Wheel cylinders belong to drum brakes and push shoes, not disc pads.
• C: Return springs pull drum shoes away from the drum when pressure drops; they do not apply disc pads.
• D (correct): Once the inner pad meets the rotor, the reaction force shifts the floating body inward so the outer pad is squeezed too.

Answer: C

Correct answer: 432 kJ. Use

All three given values must appear in the product.

• A: This is only 9 × 3.2; the temperature change of 15°C was left out of the product.
• B: This multiplies mass by temperature change but drops the specific heat.
• C (correct): Multiplying mass, specific heat, and temperature change gives 9 × 3.2 × 15 = 432 kJ.
• D: This is ten times too large, a decimal slip when multiplying by 3.2.

Answer: A

Correct answer: Chemical energy, heat, gas pressure, piston motion, rotation. Fuel holds chemical energy. Burning it releases heat, the heated gas expands and builds pressure on the piston, the piston moves in a straight line, and the connecting rod and crankshaft change that into rotary motion.

• A (correct): Burning releases the fuel's stored chemical energy as heat, the hot gas builds pressure, the pressure drives the piston straight down, and the crank turns that into rotation.
• B: This scrambles the chain; the chemical energy in the fuel comes first, and rotation is the final product, not a middle step.
• C: Rotation cannot come before the piston is pushed; the gas must be heated and pressurized first.
• D: The spark only starts the burn; the energy that drives the shaft comes from the fuel, and pressure must act before rotation occurs.

Answer: B

Correct answer: Better straight-line stability and self-centering, with heavier steering. Caster is the side-view tilt of the steering axis; with positive caster its top leans rearward, like a bicycle fork. That helps the wheels self-center and improves straight-line stability, but more of it makes steering heavier.

• A: Adding positive caster does the opposite: it strengthens the return to center after turns and adds steering effort.
• B (correct): Positive caster helps the wheels return to straight ahead and steadies the car, at the cost of extra steering effort.
• C: Shoulder wear from wheel tilt is linked to camber, the tilt seen from the front, not caster.
• D: A speed-band shake points to wheel imbalance, which caster does not change.

Answer: B

Correct answer: 19.0 m/s. Overall ratio =

so wheel speed =

Distance per minute = 1.9

m/min, and

• A: This adds the ratios (1.2 + 3.5 = 4.7) instead of multiplying them.
• B (correct): Overall ratio 4.2 gives 600 wheel rpm, and 1.9 × 600 = 1,140 m/min, or 19.0 m/s.
• C: This uses only the 1.2:1 gearbox ratio and leaves out the final-drive reduction.
• D: This is actually 1,140 meters per minute; it was not divided by 60 to get meters per second.

Answer: C

Correct answer: 0.015 Ω. The terminal voltage equals the resting voltage minus the internal drop, so first find the drop: 12.4 V minus 10.0 V is 2.4 V.

Then rearrange the drop formula

to solve for resistance.

• A: This divides the cranking voltage by the current; the internal resistance depends only on the voltage lost, not the voltage remaining.
• B: This divides the resting voltage by the current, ignoring that most of that voltage still reaches the terminals.
• C (correct): The voltage lost inside is 12.4 minus 10.0, or 2.4 V, and 2.4 V divided by 160 A is 0.015 ohm.
• D: This divides current by the voltage drop, which inverts Ohm's law.