Hydraulics and Pneumatics Lesson Test

Answer: A

Correct answer: A liquid, usually oil. Hydraulic systems transmit power through a liquid, usually oil. Because the oil hardly compresses, the motion is firm and precise and the forces can be very large.

• A (correct): Hydraulics relies on a liquid such as oil, which barely compresses and gives firm, powerful motion.
• B: A vacuum is an absence of fluid and cannot carry the high forces a hydraulic system transmits.
• C: Compressed air is the working fluid of pneumatic systems, not hydraulic ones.
• D: Steam powers steam engines and turbines; hydraulic systems push pistons with a liquid.

Answer: A

Correct answer: Force divided by area. Pressure measures how concentrated a force is:

With newtons and square meters, the unit is the pascal.

• A (correct): Pressure is P = F/A, so a given force on a smaller area gives a higher pressure.
• B: Multiplying force by area does not measure how concentrated the force is; pressure divides by area.
• C: This is the inverse of pressure and would make a smaller area give a lower value.
• D: Force times distance moved is work, not pressure.

Answer: B

Correct answer: To cap pressure by sending oil back to tank. A relief valve is held shut by a spring. When system pressure grows high enough, it opens and sends oil to the tank, setting the maximum pressure and protecting pumps, hoses and cylinders.

• A: One-way flow is the job of a check valve, not a relief valve.
• B (correct): When pressure overcomes its spring, the relief valve opens and returns oil to the reservoir, so pressure stops rising.
• C: A relief valve limits pressure for protection; it never boosts pressure.
• D: Speed is adjusted with a flow control valve that restricts flow.

Answer: D

Correct answer: A check valve. A check valve passes fluid in one direction and closes against reverse flow. In a jack, it lets each stroke add oil to the ram while preventing that oil from returning, so the load stays raised.

• A: A relief valve opens only at excess pressure to dump oil to tank; it does not hold the ram up.
• B: A regulator holds a steady downstream air pressure; it does not block reverse flow.
• C: A flow control valve limits how much oil passes per second but still allows flow both ways.
• D (correct): A check valve allows flow one way only, so oil can enter the ram cylinder but reverse flow pushes it shut.

Answer: D

Correct answer: 300 kPa. One bar equals 100 kPa, so 3 bar =

• A: This divides by 10 instead of multiplying by 100.
• B: This uses 1,000 kPa per bar, confusing the bar with the kPa-to-Pa factor.
• C: This uses 10 kPa per bar instead of 100 kPa per bar.
• D (correct): Since 1 bar equals 100 kPa, 3 bar equals 300 kPa.

Answer: A

Correct answer: A flow of oil. A pump moves oil from the reservoir into the system, so it creates flow. Pressure appears only when that flow is resisted, for example by a load on a cylinder.

• A (correct): A pump creates flow, and pressure rises only as that flow meets resistance such as a loaded cylinder.
• B: Pressure builds only when the flow meets resistance; with no load the oil circulates at low pressure.
• C: Heat is shed mainly at the reservoir and coolers, not produced as cooling by the pump.
• D: A pump converts the energy of its drive into fluid power; it cannot create energy.

Answer: D

Correct answer: A regulator. A pressure regulator keeps the pressure downstream of it at a chosen setting even when the supply pressure from the receiver changes. It cannot raise pressure above its supply.

• A: A check valve only prevents reverse flow; it does not hold pressure at a set value.
• B: A muffler quiets air leaving an exhaust port; it does not control supply pressure.
• C: A lubricator adds an oil mist for tools that need it but does not set pressure.
• D (correct): A regulator holds the downstream pressure at its setting even as the supply pressure varies.

Answer: A

Correct answer: Pass a piece of cardboard near the hose. Never use skin to find a hydraulic leak, because a pinhole jet can inject oil under the skin. Use cardboard or another tool, and depressurize the system first when possible.

• A (correct): Cardboard shows the spray without putting skin in the jet's path.
• B: Raising the relief setting can exceed part ratings and makes any jet more dangerous.
• C: A fine jet of oil can pierce skin and inject fluid under it, causing a serious injury.
• D: Ordinary gloves do not stop a high-pressure jet, so this still risks an injection injury.

Answer: B

Correct answer: Air squeezes into a smaller volume easily. Hydraulic oil barely compresses, so it gives firm, precise motion. Compressed air squeezes easily, so a pneumatic cylinder gives a little under load and its position shifts when the load changes.

• A: Air is far lighter than oil; weight is not what makes pneumatic motion springy.
• B (correct): Air is compressible, so it compresses and expands as the load changes, which makes pneumatic motion springy.
• C: Pressure travels through air lines just as it does through oil lines; that is how pneumatic tools work.
• D: Air leaks do not leave oil behind, and leakage is not the reason the motion is springy.

Answer: B

Correct answer: A hydraulic motor. Cylinders give straight-line motion, while a hydraulic motor gives continuous rotation. Pressure sets the motor's torque and flow sets how fast it spins.

• A: An accumulator stores pressurized oil for bursts of flow; it produces no motion of its own.
• B (correct): Oil flowing through a hydraulic motor turns its shaft continuously, which suits winches, tracks and conveyors.
• C: A cylinder pushes or pulls in a straight line through a fixed stroke and cannot spin a drum continuously.
• D: A relief valve limits pressure; it does not drive any output shaft.

Answer: C

Correct answer: 70 kPa.

• A: This multiplies force by area instead of dividing force by area.
• B: This is ten times too large, a decimal slip when dividing by 0.012.
• C (correct): 840 N ÷ 0.012 square meters = 70,000 Pa, which is 70 kPa.
• D: This misplaces the decimal; 840 ÷ 0.012 is 70,000 Pa, not 7,000 Pa.

Answer: A

Correct answer: 450 N. Pressure is equal on both pistons, so

• A (correct): The area ratio is 36 ÷ 4 = 9, so the output force is 50 × 9 = 450 N.
• B: This multiplies by the output area alone and ignores the input area.
• C: This multiplies by the 4 square centimeter input area instead of by the area ratio of 9.
• D: This flips the area ratio, as if the larger piston reduced the force.

Answer: C

Correct answer: 900 N. Area grows with the square of the diameter, so the area ratio is

The output force is

• A: This divides by the area ratio, as though the larger piston reduced the force.
• B: This uses the diameter ratio of 3 directly without squaring it to get the area ratio.
• C (correct): The diameter ratio is 3, so the area ratio is 3 squared, or 9, and 100 × 9 = 900 N.
• D: This cubes the diameter ratio; area depends on the square of the diameter.

Answer: D

Correct answer: 4 cm. The liquid volume is conserved:

The input displaces

cubic centimeters, so

• A: This assumes both pistons travel the same distance, but the larger piston needs more volume per centimeter.
• B: This reports the area ratio of 6 as a distance instead of dividing the travel by it.
• C: This multiplies the travel by the area ratio, which would let the output move farther than the input.
• D (correct): The input displaces 5 × 24 = 120 cubic centimeters, and 120 ÷ 30 = 4 cm.

Answer: B

Correct answer: 8 cm/s. Convert the flow: 21.6 L/min = 21,600 cubic centimeters ÷ 60 s = 360 cubic centimeters per second. Then

/s.

• A: This divides the area by the flow rate, which inverts the speed formula.
• B (correct): 21.6 L/min is 21,600 ÷ 60 = 360 cubic centimeters per second, and 360 ÷ 45 = 8 cm/s.
• C: This divides 21.6 by 45 without converting liters to cubic centimeters or minutes to seconds.
• D: This converts liters to cubic centimeters but leaves the flow per minute, so the result is really 480 cm per minute.

Answer: B

Correct answer: 36 L/min.

L/s. There are 60 seconds in a minute, so the flow is

L/min.

• A: 0.6 is the flow in liters per second; it still needs to be multiplied by 60.
• B (correct): 15 L ÷ 25 s = 0.6 L/s, and 0.6 × 60 = 36 L per minute.
• C: This multiplies volume by time instead of dividing volume by time.
• D: This divides the time by the volume, which inverts the flow-rate formula.

Answer: C

Correct answer: 360 kPa. Use Boyle's law with absolute pressures:

One third of the volume means three times the absolute pressure.

• A: This divides the pressure by 3 instead of multiplying; squeezing a gas into less space raises its pressure.
• B: This subtracts about 100 kPa of atmosphere, which would be a gauge reading, but the question asks for absolute pressure.
• C (correct): Boyle's law gives 120 times 9.0 divided by 3.0, which is 360 kPa absolute.
• D: This multiplies 120 by 9.0 but forgets to divide by the new volume of 3.0 L.

Answer: C

Correct answer: 2,000 N. Convert to SI units first: 500 kPa is 500,000 Pa, and 40 square centimeters divided by 10,000 is 0.0040 square meters.

• A: This multiplies 500 by 0.0040 without changing kilopascals to pascals, so the answer is 1,000 times too small.
• B: This converts 40 square centimeters by dividing by 100 instead of 10,000, giving an area 100 times too large.
• C (correct): Changing to 500,000 Pa and 0.0040 square meters gives 500,000 times 0.0040, which is 2,000 N.
• D: This multiplies 500 kPa by 40 square centimeters directly, mixing units, so the result is not in newtons.

Answer: A

Correct answer: 640 N. An ideal press does not create work, so input work equals output work.

The output moves one eighth as far, so it pushes eight times as hard.

• A (correct): Work in equals work out: 80 times 0.40 is 32 J, and 32 J divided by 0.05 m is 640 N.
• B: This flips the distance ratio; the piston that moves less must push harder, not more weakly.
• C: 32 is the work in joules, not the output force; it still has to be divided by the output distance.
• D: This assumes the force is unchanged, but a press trades distance for force.

Answer: D

Correct answer: 6 m/s. The same volume per second passes every section:

• A: This squares the area ratio, but speed changes in direct proportion to the area ratio.
• B: This treats the area ratio of 4 as if it were a diameter ratio and uses only its square root.
• C: This flips the area ratio; a smaller passage makes the oil speed up, not slow down.
• D (correct): By continuity, 8 × 1.5 = 2 × v, so v = 12 ÷ 2 = 6 m/s.

Answer: A

Correct answer: 49 kPa. Pressure from depth is

or 49 kPa above the surface pressure.

• A (correct): P = 1,000 × 9.8 × 5.0 = 49,000 Pa, which is 49 kPa above surface pressure.
• B: This drops a factor of 10 when converting 49,000 Pa to kilopascals.
• C: This is ten times too large, a slip when converting pascals to kilopascals.
• D: This adds atmospheric pressure, giving the absolute pressure rather than the increase caused by the water.

Answer: A

Correct answer: 9,000 N. Rod-side area =

square centimeters = 0.0045 square meters. Force = 2,000,000 Pa × 0.0045 = 9,000 N.

• A (correct): The rod-side area is 60 − 15 = 45 square centimeters, or 0.0045 square meters, and 2,000,000 Pa × 0.0045 = 9,000 N.
• B: This adds the rod area to the bore area instead of subtracting it.
• C: This uses the full bore area, which applies only when extending; on the rod side the rod covers part of the piston.
• D: This uses only the rod's area, but the oil pushes on the ring of piston around the rod.

Answer: C

Correct answer: 780 N. Convert: 15 square centimeters = 0.0015 square meters and 600 kPa = 600,000 Pa. Air force =

Net force =

• A: This adds the spring force, but the spring opposes the extending piston.
• B: This divides 15 by 100 instead of 10,000 when converting to square meters.
• C (correct): The air force is 600,000 Pa × 0.0015 square meters = 900 N, and subtracting the 120 N spring leaves 780 N.
• D: This is the full air force; it ignores the 120 N the spring takes away.

Answer: C

Correct answer: Increase the oil flow to the cylinder. Pressure times area sets force; flow divided by area sets speed. Since force is already adequate, the fix for a slow ram is more flow,

• A: A higher pressure limit adds available force, not speed, and can push parts past their ratings.
• B: A bigger piston needs more oil per centimeter, so at the same flow it moves even more slowly.
• C (correct): Ram speed equals flow divided by piston area, so more flow means faster motion.
• D: A narrower hose adds restriction and pressure drop, which tends to reduce flow and slow the ram.

Answer: D

Correct answer: 150 kPa. Pascal's principle says a pressure change in a confined fluid reaches every part of it undiminished. The pressure rises 150 kPa under both pistons; only the force is 4 times larger on the big piston.

• A: This multiplies by the square of the area ratio, mixing up area with diameter.
• B: This divides by the area ratio, as though the pressure spread thinner over the larger piston.
• C: This multiplies the pressure by the area ratio; it is the force, not the pressure, that grows with area.
• D (correct): By Pascal's principle, a pressure change in a confined fluid is passed on undiminished everywhere, including under the large piston.

Answer: C

Correct answer: An internal leak past a worn piston seal. Slow drift under load with no visible oil and a steady reservoir points to oil bypassing the piston inside the cylinder, such as past a worn seal, rather than leaking outside.

• A: Thick oil makes motion sluggish when commanded; it does not let a held load sink.
• B: An external leak would leave visible oil and lower the reservoir level, neither of which is seen.
• C (correct): Oil slipping past a worn seal inside the cylinder lets it drift under load without any visible oil loss.
• D: A high relief setting raises the maximum pressure but would not make a held load creep down.

Answer: D

Correct answer: It extends slower, but its maximum force is the same. Cylinder speed is flow rate divided by piston area,

A flow control cuts Q, so the piston moves more slowly. It does not change the piston area, and in the ideal model the force available at full pressure stays the same.

• A: Speed depends on flow, so cutting the flow does change the speed.
• B: Less oil reaching the cylinder each second cannot make it move faster.
• C: In the ideal model, restricting flow changes only speed; full pressure, and so full force, is still available when the piston stalls against a load.
• D (correct): Speed equals flow divided by piston area, so less flow means a slower piston, while the force available at full pressure is unchanged.

Answer: C

Correct answer: It can still hold pressurized oil. An accumulator stores hydraulic energy by pushing oil against compressed gas or a spring. It keeps that pressure after the pump stops, so it must be discharged as part of isolating hazardous energy.

• A: Oil thickness depends on temperature; an accumulator stores pressure and does not control viscosity.
• B: Filtering is the job of filters, not accumulators.
• C (correct): An accumulator stores oil under pressure against a gas charge or spring, so it can move parts even with the pump off.
• D: An accumulator has no control over the pump's motor; the danger is the energy it stores itself.

Answer: B

Correct answer: Stored pressure still needs isolation and controlled release. Shutting off the motor only stops new air from being added; the air already compressed in the receiver and piping keeps its energy until it is vented. That is why lockout procedures treat pressure as a separate energy source: isolate the supply, bleed the system through a controlled valve, and confirm zero pressure on a gauge before opening anything. The tempting mistake is assuming that power off means safe, but a loosened fitting can still blow out or whip a hose.

• A: Opening fittings under pressure can create serious hazards. Compressed air remains a source of stored energy despite switching off the electrical compressor.
• B (correct): Compressed gas stores energy that can cause movement or projectile hazards after electrical shutdown.
• C: A gauge can indicate pressure, and that remaining pressure is material to safety.
• D: A receiver and trapped line sections can remain under pressure after power-off.

Answer: C

Correct answer: Cold oil is thicker and resists flow. Viscosity is a fluid's thickness. Hydraulic oil is thicker when cold, so it flows less easily and actuators move sluggishly until the oil reaches operating temperature.

• A: The relief valve's spring setting does not rise with cold weather, and a higher setting would not cause slowness.
• B: Leakage past seals increases when oil is hot and thin, not when it is cold.
• C (correct): Oil viscosity rises when cold, so the pump works harder and actuators move slowly until the oil warms.
• D: Hydraulic oil barely compresses at any normal temperature.

Answer: A

Correct answer: 720 N and 2 cm. Area ratio =

Output force =

Travel shrinks by the same ratio:

Check:

J in and

J out.

• A (correct): The area ratio is (8/2) squared = 16, so force is 45 × 16 = 720 N and travel is 32 ÷ 16 = 2 cm.
• B: This flips the area ratio, reducing force and multiplying travel, the opposite of what a larger output piston does.
• C: The force correctly uses the area ratio, but the travel uses only the diameter ratio, so work out would exceed work in.
• D: This uses the diameter ratio of 4 instead of the area ratio of 16 for both force and travel.

Answer: D

Correct answer: 15.7 kN.

or about 15.7 kN.

• A: This uses the diameter squared as the area and leaves out the factor of pi over 4.
• B: This divides square centimeters by 100 instead of 10,000 when converting to square meters.
• C: This uses the 5.0 cm diameter as if it were the radius, making the area four times too large.
• D (correct): Area = pi × 0.050 squared ÷ 4 ≈ 0.00196 square meters, and 8,000,000 Pa × 0.00196 ≈ 15,700 N.

Answer: D

Correct answer: Pressure climbs to the relief setting, and oil returns to tank through the relief valve. A positive-displacement pump delivers flow whatever the load. When the cylinder bottoms out, the flow has nowhere to go, so pressure rises until the relief valve opens and sends the oil back to the tank, protecting the circuit.

• A: A positive-displacement pump traps and forces out a fixed volume each turn regardless of the load.
• B: Blocked flow meets maximum resistance, so pressure rises rather than falls.
• C: The cylinder is already at the end of its stroke, so no extra oil can enter it to move the piston.
• D (correct): With the outlet blocked, the pump keeps forcing oil, pressure rises until the relief valve opens, and the oil goes back to the reservoir.

Answer: B

Correct answer: 540 kPa. Change the gauge reading to absolute:

Apply Boyle's law:

The gauge reads

• A: This applies Boyle's law to the gauge reading directly; gas-law calculations need absolute pressure.
• B (correct): Absolute pressure goes from 400 kPa to 400 times 8.0 divided by 5.0, which is 640 kPa, and the gauge shows 640 minus 100, or 540 kPa.
• C: 640 kPa is the new absolute pressure; a gauge reads about 100 kPa less than that.
• D: This uses the volume ratio upside down, which would lower the pressure even though the air is being squeezed.

Answer: B

Correct answer: 8 cm/s. On the rod side the effective area is

square centimeters. Speed = flow ÷ area =

/s, faster than the 6 cm/s extending speed.

• A: This divides by the rod's area, but the oil acts on the ring of piston around the rod.
• B (correct): The rod-side area is 40 − 10 = 30 square centimeters, and 240 ÷ 30 = 8 cm/s.
• C: This adds the rod area to the bore area instead of subtracting it.
• D: This divides by the full bore area, which gives the extending speed, not the retracting speed.

Answer: C

Correct answer: 12 strokes. Volume per stroke =

cubic centimeters. Volume to raise the ram 3 cm =

cubic centimeters. Strokes =

• A: This divides the 120 cubic centimeters needed by the area ratio instead of by the 10 cubic centimeters delivered per stroke.
• B: This is just the area ratio; it does not account for the lift height or the stroke volume.
• C (correct): Each stroke moves 2 × 5 = 10 cubic centimeters, the ram needs 40 × 3 = 120 cubic centimeters, and 120 ÷ 10 = 12.
• D: This multiplies the 3 cm lift by the area ratio of 20 but ignores the 5 cm the pump piston moves on each stroke.

Answer: D

Correct answer: 230 kPa. Convert to absolute:

absolute. The gauge reads

• A: This is the new absolute pressure; the gauge reads 100 kPa less.
• B: This applies the temperature ratio to the gauge pressure, but gas laws require absolute pressure.
• C: This uses Celsius temperatures (57 over 27) instead of kelvin and also skips the absolute-pressure conversion.
• D (correct): Absolute pressure goes from 300 kPa to 300 × 330 ÷ 300 = 330 kPa, and subtracting 100 kPa gives a gauge reading of 230 kPa.

Answer: B

Correct answer: 700 kPa. The air force must equal load plus spring:

With 20 square centimeters = 0.0020 square meters,

Pa = 700 kPa.

• A: This divides 20 by 100 instead of 10,000 when converting the area to square meters.
• B (correct): The air must supply 1,200 + 200 = 1,400 N, and 1,400 N ÷ 0.0020 square meters = 700,000 Pa.
• C: This is enough for the 1,200 N load alone but leaves nothing to overcome the spring.
• D: This subtracts the spring force, but the spring works against the extending piston and must be added.

Answer: A

Correct answer: Use a cylinder with a larger piston area. Since

and the pressure is already at its rated limit, the safe way to gain force is a larger piston area, or a system designed for higher pressure. Overriding ratings risks a dangerous failure.

• A (correct): Force equals pressure times area, so a bigger piston gives more force without exceeding the pressure rating.
• B: A narrower hose adds restriction and pressure loss; it cannot increase the force at the piston.
• C: This pushes hoses, fittings and the cylinder past their rated pressure, risking a sudden burst.
• D: More flow makes the cylinder move faster but does not add force at the same pressure.

Answer: B

Correct answer: 12 s. Volume needed:

cubic centimeters. Flow: 9,000 cubic centimeters per minute divided by 60 is 150 cubic centimeters per second.

• A: This divides 1,800 cubic centimeters by 15 instead of 150, a decimal slip in the flow rate.
• B (correct): The stroke needs 1,800 cubic centimeters, and 9 L per minute is 150 cubic centimeters per second, so it takes 12 s.
• C: 0.2 minute was turned into seconds by multiplying by 100 instead of 60.
• D: This divides 1.8 L by 9 L correctly but reads the result as seconds; the flow rate is per minute, so the answer is 0.2 minute.

Answer: A

Correct answer: 480 kPa. Convert the area first:

• A (correct): 25 square centimeters is 0.0025 square meters, and 1,200/0.0025 = 480,000 Pa, or 480 kPa.
• B: This converts square centimeters to square meters by dividing by 100 instead of 10,000.
• C: This divides newtons by square centimeters without converting the area to square meters.
• D: This converts the area with one too many zeros, using 0.00025 square meters instead of 0.0025.

Answer: D

Correct answer: 5.0 L. Boyle's law with absolute pressures:

The pressure rose 2.4 times, so the volume fell to 1/2.4 of its starting value.

• A: This inverts the pressure ratio; raising the pressure must shrink the volume, not enlarge it.
• B: This converts both pressures to gauge values first; Boyle's law requires absolute pressures.
• C: This assumes the volume stays fixed; at constant temperature pressure and volume are inversely related.
• D (correct): By Boyle's law, the new volume is 150 times 12.0 / 360 = 5.0 L; higher pressure means smaller volume.

Answer: B

Correct answer: 0.75 L. Displacement is

• A: This divides 750 cubic centimeters by 100 instead of 1,000 when converting to liters.
• B (correct): Volume is area times stroke: 30 times 25 = 750 cubic centimeters, which is 0.75 L.
• C: This divides 750 cubic centimeters by 10,000; one liter is 1,000 cubic centimeters.
• D: This divides the area by the stroke; displacement is area multiplied by stroke length.

Answer: C

Correct answer: 12.6 s. Find the piston area from the diameter, then the volume, then divide by the flow rate.

The flow is

cubic centimeters per second, so

• A: This squares the diameter but leaves out the $\pi /4$ factor, overstating the area as 64 square centimeters.
• B: This uses $\pi d^2$ without dividing by 4, making the area and the oil volume four times too large.
• C (correct): Piston area is $\pi (8.0{)}^2/4\approx 50.3$ square centimeters, so the stroke takes about 2,513 cubic centimeters; at 200 cubic centimeters per second that is about 12.6 s.
• D: This uses the 8.0 cm diameter as if it were the area; area grows with the square of the diameter.

Answer: B

Correct answer: The filter. An air preparation unit is often called an FRL: the filter removes particles and liquid water, the regulator sets the working pressure, and the lubricator adds oil mist where tools need it.

• A: The regulator sets a steady downstream pressure; it does not clean the air.
• B (correct): The filter traps particles and separates out water before the air reaches tools and valves.
• C: The lubricator adds a fine oil mist for tools; it adds something rather than removing contaminants.
• D: A muffler quiets air leaving an exhaust port; it does not treat the supply air.

Answer: C

Correct answer: 250 N and 24 cm. Square the diameter ratio to get the area ratio:

Force shrinks by the area ratio and travel grows by the same ratio, so work in equals work out.

• A: This uses the diameter ratio of 4 for both force and travel; area, which sets both, grows with the square of the diameter.
• B: The force correctly uses the area ratio of 16, but the travel uses the diameter ratio of 4 instead of the same area ratio.
• C (correct): The area ratio is (6.0/1.5) squared = 16, so the input force is 4,000/16 = 250 N and the input travel is 16 times 1.5 = 24 cm.
• D: The travel correctly uses the area ratio of 16, but the force uses the unsquared diameter ratio of 4.

Answer: A

Correct answer: The working fluid is air. Pneumatic and hydraulic symbols look much alike, but pneumatic triangles are drawn hollow to show air, while hydraulic triangles are filled to show liquid.

• A (correct): Hollow triangles mark pneumatic components; filled triangles mark hydraulic, liquid-filled ones.
• B: Oil, a liquid, is shown with filled triangles, not hollow ones.
• C: Triangles show the direction of energy flow by where they point, not by whether they are filled.
• D: A return to tank is shown by a line ending at the reservoir symbol, not by hollow triangles.

Answer: C

Correct answer: 35.3 kPa. Pressure below the surface of a still liquid rises by $\rho gh$.

• A: This uses the density of water, 1,000 kg per cubic meter, instead of the oil's 900.
• B: This multiplies density by depth but leaves out g, which gives 3,600 instead of 35,280 Pa.
• C (correct): Hydrostatic pressure is $\rho gh=900\times 9.8\times 4.0=35,280$ Pa, about 35.3 kPa.
• D: This adds atmospheric pressure, but the question asks only how much greater the bottom pressure is than the surface.

Answer: D

Correct answer: 1.8 m/s. Convert flow to cubic centimeters per second, then use

• A: This converts 180 cm/s to meters per second by dividing by 1,000 instead of 100.
• B: This converts 180 cm/s by dividing by 10 instead of 100, a factor-of-ten unit slip.
• C: This skips converting minutes to seconds, so the value is really 108 meters per minute, not per second.
• D (correct): 27 L/min is 27,000/60 = 450 cubic centimeters per second; 450/2.5 = 180 cm/s, which is 1.8 m/s.

Answer: D

Correct answer: Air trapped in the brake lines. Brake hydraulics rely on fluid that barely compresses, which gives a firm pedal. Opening a line can let air in, and because air squeezes easily, pressing the pedal compresses the bubbles first, so the pedal feels spongy and braking is weaker until the lines are bled.

• A: Barely compressible fluid is what makes a healthy pedal feel firm, not spongy.
• B: Larger wheel pistons are the normal design for multiplying force; they do not cause a spongy feel.
• C: A relief setting limits maximum pressure; it would not make the pedal feel soft at light pressure.
• D (correct): Air compresses easily, so pedal travel is used squeezing the air bubbles instead of moving the wheel pistons.