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EASA CPL(A) technical and operational foundations study materials
EASA CPL(A) technical and operational foundations study resources

EASA CPL(A) technical and operational foundations practice questions

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Question 1Operational framework and communications

Under the Chicago Convention of 1944, which Article establishes that every contracting State has complete and exclusive sovereignty over the airspace above its territory?

  1. Article 1 Correct answer.
  2. Article 5
  3. Article 12
  4. Article 33

Explanation

Article 1 of the Convention on International Civil Aviation (Chicago Convention) explicitly recognises that every State has complete and exclusive sovereignty over the airspace above its territory. This principle is the cornerstone of international air law.

Question 2Operational framework and communications

When two power-driven heavier-than-air aeroplanes are converging at approximately the same altitude on crossing courses, which rules of the air apply under SERA.3210?

  1. The aircraft that has the other on its right shall give way Correct answer.
  2. The aircraft that has the other on its left shall give way
  3. Both aircraft shall turn to their respective left immediately
  4. The faster aircraft must climb while the slower aircraft maintains course

Explanation

Under SERA.3210(c)(1), when two aircraft of the same category are converging at approximately the same level, the aircraft that has the other on its right shall give way by altering course or avoiding passing over, under, or ahead.

Question 3Operational framework and communications

If an aeronautical station cannot establish two-way communication but has reason to believe the aircraft can receive, how should a blind transmission be made?

  1. Transmit the message twice on the designated frequency, preceded by the phrase TRANSMITTING BLIND Correct answer.
  2. Transmit the message once on the assigned frequency, preceded by the phrase RADIO CHECK
  3. Repeat each word twice, preceded by the phrase WORDS TWICE
  4. Move immediately to 121.500 MHz and send a distress relay regardless of the message priority

Explanation

A station making a blind transmission uses the phrase TRANSMITTING BLIND and transmits the message twice on the designated frequency. Other available frequencies on which the aircraft might be listening may also be used when appropriate.

Question 4Operational framework and communications

Under ICAO Annex 10 and EASA radiotelephony rules, what is the correct standard phraseology for a pilot to acknowledge receipt of an ATC message containing no actionable clearance or readback requirement?

  1. ROGER followed by the aircraft callsign Correct answer.
  2. OVER AND OUT
  3. AFFIRMATIVE ROGER
  4. COPY THAT OVER

Explanation

ROGER means that all of the last transmission has been received and is appropriate when there is nothing to comply with or read back. WILCO additionally means that the message is understood and will be complied with, so it is not the precise response to a purely informational transmission.

Question 5Operational framework and communications

What is the international distress signal spoken on VHF radiotelephony, and what priority does it have over all other communications?

  1. 'MAYDAY', spoken three times; it has absolute priority over all other communications Correct answer.
  2. 'PAN PAN', spoken three times; it has equal priority with commercial scheduled traffic
  3. 'MAYDAY RELAY', spoken three times; it is used by the aircraft in distress as the normal opening of its own call
  4. 'PAN PAN', spoken once; it has priority over distress traffic when no assistance is requested

Explanation

'MAYDAY' (from French m'aider) repeated three times indicates an aircraft is threatened by grave and imminent danger and requests immediate assistance. Distress messages have absolute priority over all other transmissions.

Question 6Operational framework and communications

Which VHF frequency should an aircraft monitor when the SERA emergency-frequency listening-watch requirement applies?

  1. 121.500 MHz Correct answer.
  2. 123.450 MHz
  3. 118.100 MHz
  4. 122.800 MHz

Explanation

121.500 MHz is the international VHF aeronautical emergency frequency. Aircraft should maintain a watch when practicable in the specifically prescribed circumstances, including long over-water flights and flights over designated areas where an emergency locator transmitter may be required, except while communications on other VHF channels or equipment limitations prevent a simultaneous watch.

Question 7Operational framework and communications

If an aircraft experiences complete communications transmitter failure but can still receive, what procedure should the pilot execute when receiving an ATC transmission?

  1. Acknowledge transmissions by squawking ident or changing transponder code as directed by ATC Correct answer.
  2. Set code 7600 and disregard any ATC request to acknowledge by IDENT or code change
  3. Change frequency after every received transmission to show that the receiver remains serviceable
  4. Use the speechless-transmission procedure and key the microphone without modulation

Explanation

When an aircraft has a transmitter failure but receiver functioning, the pilot sets transponder code 7600 and monitors the appropriate ATC frequency. ATC will instruct the pilot to 'squawk ident' or make specific turns or squawk changes to acknowledge instructions.

Question 8Operational framework and communications

On an Automatic Terminal Information Service (ATIS) broadcast, how does a pilot confirm that they have received the most current airfield information when first contacting ATC?

  1. By stating the phonetic alphabet identification letter of the current ATIS broadcast (e.g. 'Information Charlie received') Correct answer.
  2. By reading back the QNH and runway only, because those two values uniquely identify each ATIS edition
  3. By stating the UTC observation time, because the information letter is optional
  4. By saying 'ATIS received' without its identification letter unless ATC asks for the letter

Explanation

Each sequential ATIS broadcast is identified by a phonetic letter from the ICAO alphabet (Information Alpha, Bravo, Charlie, etc.). Pilots must state the current information designator on initial contact (e.g. 'Copenhagen Approach, OY-ABC, passing 3,000 ft, Information Delta received, QNH 1012') to confirm they hold the latest weather and operational notices.

Question 9Aircraft systems and instruments

Why can a serviceable magneto-ignition piston engine continue running after the aircraft battery master switch is turned off?

  1. Each engine-driven magneto generates its own ignition current independently of the aircraft electrical system Correct answer.
  2. The battery feeds conventional ignition coils through a circuit that bypasses the master contactor
  3. The alternator supplies low-voltage current directly to the spark plugs after engine start
  4. The starter motor remains energised at low power and drives the ignition distributor

Explanation

A magneto is an engine-driven, self-contained electrical generator for the ignition system. Once the engine is turning, it does not need the aircraft battery or alternator to energise the spark plugs; the magneto switch and its earthing leads control whether it operates.

Question 10Aircraft systems and instruments

In a modern semi-monocoque aircraft fuselage structure, which structural members primarily absorb and carry the longitudinal bending loads?

  1. Longerons and stringers Correct answer.
  2. Formers and bulkheads
  3. Skin panels exclusively
  4. Floor beams and seat rails

Explanation

In semi-monocoque construction, longerons and stringers run longitudinally to carry tensile and compressive bending loads, while the thin skin resists shear and torsional stress, and formers/bulkheads provide cross-sectional shape and rigidity.

Question 11Aircraft systems and instruments

What is the key functional difference between a lead-acid aircraft battery and a nickel-cadmium (NiCd) battery during high-rate discharge?

  1. A NiCd battery maintains an almost constant voltage until near complete discharge, but is susceptible to thermal runaway if improperly charged Correct answer.
  2. A lead-acid battery maintains the flatter discharge voltage, while NiCd voltage falls steadily with remaining capacity
  3. A NiCd battery voltage drops linearly and proportionally with state of charge, providing clear early warning of discharge
  4. The two chemistries have essentially identical voltage-discharge behaviour and charging hazards

Explanation

NiCd batteries have low internal resistance and maintain a stable terminal voltage throughout their discharge cycle, but excessive charge currents at high cell temperatures can trigger thermal runaway. Lead-acid batteries display a steady voltage drop and their electrolyte freezing point rises when discharged.

Question 12Aircraft systems and instruments

What is the primary danger associated with severe detonation in an aircraft reciprocating piston engine?

  1. Extremely rapid pressure rise and explosive shock waves that erode piston crowns, burn valves, and can cause sudden structural cylinder failure Correct answer.
  2. A slow, controlled flame front that merely reduces combustion efficiency without abnormal pressure peaks
  3. Spark-plug fouling from a rich mixture, with no damaging pressure or temperature rise
  4. Pre-ignition occurring only after the normal spark and producing no shock-wave loading

Explanation

Detonation is the instantaneous, uncontrolled explosion of the unburned fuel-air end-gas ahead of the flame front. It creates extreme acoustic shock waves and dramatic local temperature spikes that can rapidly melt pistons, crack cylinder heads, and destroy the engine.

Question 13Aircraft systems and instruments

In an aircraft fuel system, why are fuel tank vent lines equipped with anti-ice designs or backward-facing relief vents?

  1. To prevent tank structural collapse or fuel starvation if ice blocks the main vent openings during flight Correct answer.
  2. To maintain a high positive tank pressure as the normal source of fuel flow to the engine
  3. To accommodate thermal expansion only; a blocked vent cannot affect in-flight fuel delivery
  4. To return excess metered fuel from the carburettor to the tank

Explanation

If fuel tank vents freeze over, the continuous consumption of fuel creates a vacuum within the tank. This can lead to fuel starvation (stopping the engine) or structural distortion/collapse of the wing skin. Backup or anti-ice vents ensure atmospheric pressure is maintained.

Question 14Aircraft systems and instruments

What is the function of the bypass valve located in a full-flow aircraft engine oil lubrication filter assembly?

  1. To ensure continued engine lubrication with unfiltered oil if the filter element becomes clogged with contaminants Correct answer.
  2. To divert hot oil directly to the oil cooler whenever oil temperature exceeds 100°C
  3. To maintain constant oil pressure at the propeller governor regardless of engine RPM
  4. To route oil back to the sump whenever the main oil-pressure relief valve opens

Explanation

If an oil filter element becomes clogged with carbon or metal debris, the differential pressure across the element rises until the spring-loaded bypass valve opens. This ensures the engine continues to receive lubricating oil (unfiltered) rather than suffering immediate oil starvation and catastrophic seizure.

Question 15Aircraft systems and instruments

If the pitot inlet becomes completely blocked by ice while the pitot drain hole and the static source remain open, what indication should the pilot expect?

  1. The ASI will fall toward zero because the pitot line vents through the open drain Correct answer.
  2. The ASI will act like an altimeter, falsely indicating an increase in airspeed as altitude increases
  3. The ASI will freeze at the exact speed at which the blockage occurred regardless of climb or descent
  4. The ASI will continue to indicate normally because the open static source supplies all pressure information it needs

Explanation

With the pitot inlet blocked but the drain open, the pitot line vents and cannot retain total pressure. Pitot and static pressure at the ASI then become approximately equal, so the indicated airspeed falls toward zero rather than changing progressively like an altimeter.

Question 16Aircraft systems and instruments

In an aircraft turn coordinator, what specific motion of the aircraft does the miniature aircraft symbol depict, and how does it differ from a traditional turn and slip indicator?

  1. The miniature aircraft senses both roll rate and yaw rate because its gyro gimbal is tilted upward by approximately 30° Correct answer.
  2. The miniature aircraft senses only yaw rate because its gyro spin axis is aligned purely vertically
  3. The miniature aircraft displays pitch angle and bank angle directly via a 3-axis pendulous gimbal
  4. The miniature aircraft displays sideslip angle derived from differential pitot pressures

Explanation

In a turn coordinator, the gyro gimbal frame is canted upward at an angle of roughly 30° from the longitudinal axis. This cant allows the gyro to respond to both roll rate (during turn entry/exit) and yaw rate, giving faster response than a turn and slip indicator (which senses yaw only).

Question 17Aircraft systems and instruments

What is the cause of apparent wander (apparent drift) in a directional gyroscope operating on the Earth's surface?

  1. The Earth's rotation beneath the gyroscopic spin axis, which remains fixed in inertial space Correct answer.
  2. Bearing friction and dynamic imbalance within the gimbal rings
  3. Variations in local magnetic anomalies and terrestrial iron deposits
  4. Centrifugal forces acting on the gyro rotor during coordinated turns

Explanation

Apparent wander occurs because a free gyroscope maintains its axis pointing in a fixed direction relative to inertial space (gyroscopic rigidity), while the Earth rotates underneath it at 15° per hour multiplied by the sine of the latitude.

Question 18Aircraft systems and instruments

When flying in the Northern Hemisphere on an easterly or westerly heading, what error does the direct-reading magnetic compass display when the aircraft accelerates?

  1. The compass indicates a false turn toward the North Correct answer.
  2. The compass indicates a false turn toward the South
  3. The compass indicates correctly without any lag or acceleration error
  4. The compass initially lags the true heading but shows no northerly or southerly deflection

Explanation

Under the ANDS rule (Accelerate North, Decelerate South) in the Northern Hemisphere, inertia causes the compass card's center of gravity (which is pendulously suspended and offset) to lag, causing the card to rotate and falsely indicate a turn toward the North during acceleration.

Question 19Aircraft systems and instruments

What is the primary function of an Air Data Computer (ADC) in an integrated glass cockpit / EFIS avionics suite?

  1. To receive raw pitot, static, and total temperature inputs and compute calibrated airspeed, true airspeed, Mach number, and pressure altitude Correct answer.
  2. To calculate GPS satellite pseudo-ranges and verify receiver autonomous integrity monitoring (RAIM)
  3. To compute engine cylinder head temperatures and control automatic mixture scheduling
  4. To generate artificial horizon pitch and roll attitude references using solid-state ring laser gyros

Explanation

An Air Data Computer (ADC) samples total pressure from the pitot tube, static pressure from static ports, and outside air temperature (OAT) to compute air data parameters including CAS, TAS, Mach number, altitude, and vertical speed for display on the PFD.

Question 20Aircraft systems and instruments

On an airspeed indicator, what aerodynamic limit is defined by the upper boundary of the green operating arc?

  1. Vno - Maximum structural cruising speed Correct answer.
  2. Vne - Never-exceed speed
  3. Vfe - Maximum flap extended speed
  4. Va - Design maneuvering speed

Explanation

The green arc on an ASI indicates the normal operating range. The lower limit is Vs1 (clean stall speed) and the upper limit is Vno (maximum structural cruising speed). Flying above Vno is only permitted in smooth air.

Question 21Aircraft systems and instruments

What happens to the mixture and engine power when carburettor heat is selected on a normally aspirated piston engine without automatic mixture compensation?

  1. The warmer, less-dense intake air makes the mixture richer and normally reduces available power Correct answer.
  2. The warmer intake air makes the mixture leaner and normally increases available power
  3. The mixture remains unchanged because the carburettor automatically compensates for the lower air density
  4. The warmer air makes the mixture leaner even though available power decreases

Explanation

Carburettor heat admits air that has been warmed and is therefore less dense. With a fixed fuel metering setting, less air mass combines with approximately the same fuel flow, enriching the mixture; the lower charge density also normally reduces power.

Question 22Aircraft systems and instruments

A 24-volt aircraft battery supplies a steady 12-ampere load for 90 minutes. Ignoring capacity-rate and temperature effects, how much battery capacity is consumed?

  1. 18 ampere-hours Correct answer.
  2. 8 ampere-hours
  3. 36 ampere-hours
  4. 432 ampere-hours

Explanation

Capacity used is current multiplied by time in hours. Ninety minutes is 1.5 hours, so 12 A × 1.5 h = 18 Ah. The 24-volt rating is relevant to system voltage but is not multiplied into an ampere-hour calculation.

Question 23Aircraft systems and instruments

Why is vapour lock more likely in an aircraft fuel line at high altitude or with unusually warm fuel?

  1. Lower ambient pressure and higher fuel temperature both make vaporisation in the line more likely, which can interrupt liquid fuel flow Correct answer.
  2. Higher altitude increases fuel density until the fuel can no longer pass through the filter
  3. Warm fuel raises ambient pressure around the line and forces the mixture excessively lean
  4. Vapour lock is caused only by water freezing in the tank vents

Explanation

Fuel boils more readily as pressure decreases and temperature rises. Vapour bubbles in a suction-side fuel line can reduce or stop delivery to the carburettor or injection system, so routing, boost pumps and fuel-system design are used to preserve adequate pressure.

Question 24Aircraft systems and instruments

Why can a dark area immediately beyond a strong weather-radar return be hazardous rather than reassuring?

  1. Heavy precipitation may attenuate the transmitted and returned energy, creating a radar shadow that conceals weather behind it Correct answer.
  2. The antenna tilt may place the beam above low-level precipitation at long range, so every dark area behind a return is confirmed clear
  3. Automatic gain control removes all weak echoes behind a strong cell but preserves a reliable weather-free indication
  4. Ground clutter masks the strong return itself, while the region beyond it remains fully surveyed

Explanation

Very heavy precipitation can absorb and scatter so much radar energy that little reaches or returns from cells farther away. The resulting dark radar shadow can therefore indicate attenuation behind a strong cell, not an absence of hazardous weather.

Question 25Aircraft systems and instruments

In a pressurised aeroplane, how does the outflow valve normally control cabin pressure?

  1. It meters the discharge of cabin air while the compressors continue supplying inflow Correct answer.
  2. It meters incoming bleed-air flow while keeping the cabin discharge opening fixed
  3. It acts only as a positive-pressure relief valve and remains fully closed during normal regulation
  4. It changes compressor speed to hold a constant cabin inflow and seals all normal discharge paths

Explanation

A continuous source, commonly engine bleed air, supplies conditioned air to the cabin. The pressure controller positions the outflow valve to regulate how quickly that air escapes, thereby maintaining the scheduled cabin altitude and differential pressure.

Question 26Performance, loading and principles of flight

An aeroplane has an empty mass of 800 kg with a center of gravity located at station 2.10 m aft of the datum. The pilot and front passenger (total mass 160 kg) sit at station 2.30 m. Fuel loaded is 100 kg at station 2.50 m. What is the total mass and the resulting position of the center of gravity?

  1. Total mass = 1,060 kg; CG = 2.168 m aft of datum Correct answer.
  2. Total mass = 1,060 kg; CG = 2.245 m aft of datum
  3. Total mass = 1,000 kg; CG = 2.120 m aft of datum
  4. Total mass = 1,060 kg; CG = 2.050 m aft of datum

Explanation

Calculate total moments: Empty mass: 800 * 2.10 = 1,680 kg·m. Front crew: 160 * 2.30 = 368 kg·m. Fuel: 100 * 2.50 = 250 kg·m. Total mass = 800 + 160 + 100 = 1,060 kg. Total moment = 1,680 + 368 + 250 = 2,298 kg·m. CG = Total moment / Total mass = 2,298 / 1,060 ≈ 2.1679 m (2.168 m).

Question 27Performance, loading and principles of flight

What aerodynamic consequence occurs when an aeroplane is loaded such that its center of gravity is located aft of the aft CG certified limit?

  1. Longitudinal stability decreases dramatically, control stick forces become dangerously light, and stall/spin recovery may become impossible Correct answer.
  2. Longitudinal stability increases dramatically, making the aircraft nose-heavy and requiring excessive back pressure to flare
  3. The stalling speed increases markedly due to increased tail downforce requirements
  4. Cruising range decreases substantially due to high trim drag caused by excessive downward elevator deflection

Explanation

An aft CG reduces the distance between the CG and the aerodynamic center (shortening the static margin). This severely degrades longitudinal static and dynamic stability, lightens elevator control forces, and can prevent the elevator from pitching the nose down to recover from a stall.

Question 28Performance, loading and principles of flight

An aircraft with a total mass of 1,200 kg has its CG located at station 2.40 m. A 50 kg baggage box is shifted aft from station 1.80 m to station 3.00 m. By how much does the CG shift, and what is its new location?

  1. CG shifts aft by 0.050 m; new CG = station 2.45 m Correct answer.
  2. CG shifts aft by 0.025 m; new CG = station 2.425 m
  3. CG shifts forward by 0.050 m; new CG = station 2.35 m
  4. CG shifts aft by 0.120 m; new CG = station 2.52 m

Explanation

Using the mass shift formula: ΔCG = (Mass shifted * Distance shifted) / Total mass. Distance shifted = 3.00 - 1.80 = 1.20 m aft. ΔCG = (50 * 1.20) / 1,200 = 60 / 1,200 = 0.050 m aft. New CG = 2.40 + 0.050 = 2.45 m.

Question 29Performance, loading and principles of flight

In mass and balance calculations, what does the term 'Mean Aerodynamic Chord' (MAC) represent, and how is the CG position usually expressed relative to it in larger aeroplanes?

  1. The chord of an equivalent rectangular wing having identical aerodynamic characteristics; CG is expressed as a percentage of MAC (% MAC) aft of the leading edge of the MAC (LEMAC) Correct answer.
  2. The physical geometric chord measured at the aircraft fuselage centerline; CG is expressed in meters forward of the trailing edge
  3. The width of the horizontal stabilizer; CG is expressed as a percentage of rudder deflection angle
  4. The distance between the nose landing gear and the main landing gear axles expressed as an index unit

Explanation

The Mean Aerodynamic Chord (MAC) is the chord of an imaginary untwisted rectangular wing that has the same lift and pitching moment characteristics as the actual wing. CG position is standardly defined as % MAC measured aft of the Leading Edge of MAC (LEMAC).

Question 30Performance, loading and principles of flight

An aerodrome has a pressure altitude of 4,000 ft and an outside air temperature (OAT) of +27°C. Using the standard atmospheric lapse rate rule of thumb, what is the approximate density altitude at this aerodrome?

  1. 6,400 ft Correct answer.
  2. 4,000 ft
  3. 5,200 ft
  4. 7,800 ft

Explanation

Standard temperature at sea level = +15°C. At 4,000 ft, ISA temperature = 15 - (2 * 4) = +7°C. Actual temperature = +27°C, which is ISA + 20°C. Density Altitude = Pressure Altitude + (120 * ISA deviation) = 4,000 + (120 * 20) = 4,000 + 2,400 = 6,400 ft.

Question 31Performance, loading and principles of flight

How does an uphill runway slope affect the take-off and landing distances of an aeroplane?

  1. An uphill slope increases the take-off ground run and decreases the landing roll Correct answer.
  2. An uphill slope decreases the take-off ground run and increases the landing roll
  3. An uphill slope increases both the take-off ground run and the landing roll
  4. An uphill slope has zero aerodynamic or physical effect on wheel-braked landing distance

Explanation

A component of gravity acts opposite to the direction of motion when accelerating uphill, reducing net acceleration and lengthening the take-off run. Conversely, when landing uphill, gravity acts in the same direction as braking force, decelerating the aircraft faster and shortening landing roll.

Question 32Performance, loading and principles of flight

What is the aerodynamic definition of the best rate of climb speed (Vy) in an unsupercharged piston aeroplane?

  1. The calibrated airspeed where the maximum excess engine power (Power Available minus Power Required) occurs Correct answer.
  2. The calibrated airspeed where the maximum excess thrust (Thrust Available minus Thrust Required) occurs
  3. The airspeed that achieves the steepest climb angle per unit horizontal ground distance
  4. The minimum controllable airspeed in the clean configuration with maximum continuous power

Explanation

Best rate of climb (Vy) produces the greatest gain in altitude per unit of time and occurs at the airspeed where the maximum excess power (Power Available - Power Required) exists. Best angle of climb (Vx) occurs at maximum excess thrust.

Question 33Performance, loading and principles of flight

In a twin-engine light aeroplane, what is the aerodynamic effect on the one-engine-inoperative (OEI) minimum control speed (Vmca) if the center of gravity is moved forward?

  1. Vmca decreases because the increased rudder moment arm generates a larger yaw-correcting moment Correct answer.
  2. Vmca increases because the rudder moment arm is shortened, reducing directional authority
  3. Vmca remains unchanged because Vmca depends exclusively on asymmetric engine thrust and air density
  4. Vmca increases because the stalling speed increases with a forward CG

Explanation

A forward CG increases the distance between the center of gravity and the vertical tail/rudder. This longer moment arm enables the rudder to generate a greater yaw-countering moment for a given deflection, allowing directional control to be maintained down to a lower airspeed (decreasing Vmca).

Question 34Performance, loading and principles of flight

What is the crosswind component on Runway 04 (magnetic heading 040°) when the surface wind is reported as 080°/20 kt?

  1. 12.9 kt Correct answer.
  2. 15.3 kt
  3. 10.0 kt
  4. 18.8 kt

Explanation

Angle between runway heading and wind direction = 080° - 040° = 40°. Crosswind component = Wind speed * sin(40°) = 20 * 0.6428 ≈ 12.86 kt (rounded to 12.9 kt).

Question 35Performance, loading and principles of flight

What happens to the absolute ceiling and service ceiling of an unsupercharged piston-engine aeroplane when gross mass is reduced?

  1. Both the absolute ceiling and service ceiling increase to higher altitudes Correct answer.
  2. Both ceilings decrease because the propeller produces less thrust at lower mass
  3. The service ceiling increases but the absolute ceiling remains completely unchanged
  4. The absolute ceiling decreases while the service ceiling increases

Explanation

A lower aircraft mass reduces the lift required to support the aircraft, which decreases induced drag and reduces the power required for level flight. Consequently, excess power exists up to higher altitudes, increasing both the service ceiling (100 ft/min climb capability) and absolute ceiling (zero climb rate).

Question 36Performance, loading and principles of flight

In steady, unaccelerated level flight, an aeroplane has a total mass of 1,200 kg. If the pilot enters a coordinated, constant-altitude turn at a bank angle of 60°, what is the resulting load factor (n) and apparent weight supported by the wings?

  1. Load factor = 2.0; required lift = approximately 23,544 N Correct answer.
  2. Load factor = 1.41; required lift = approximately 16,600 N
  3. Load factor = 1.15; required lift = approximately 13,540 N
  4. Load factor = 3.0; required lift = approximately 35,316 N

Explanation

In a coordinated constant-altitude turn, n = 1 / cos(bank angle). At 60°, n = 1 / 0.5 = 2.0. Aircraft weight is 1,200 × 9.81 = 11,772 N, so the lift required is 2 × 11,772 = 23,544 N.

Question 37Performance, loading and principles of flight

If an aeroplane has a clean, level-flight stalling speed (Vs1) of 60 kt CAS, what will its stalling speed be in a coordinated 60° bank level turn?

  1. 85 kt CAS Correct answer.
  2. 120 kt CAS
  3. 70 kt CAS
  4. 60 kt CAS

Explanation

Stalling speed in a turn increases with the square root of the load factor: Vs_turn = Vs * sqrt(n). At 60° bank, n = 2.0. Vs_turn = 60 * sqrt(2) ≈ 60 * 1.414 = 84.85 kt (round to 85 kt CAS).

Question 38Performance, loading and principles of flight

How does induced drag vary with changes in aeroplane airspeed in unaccelerated level flight?

  1. Induced drag is inversely proportional to the square of the airspeed (decreases as airspeed increases) Correct answer.
  2. Induced drag is directly proportional to the square of the airspeed (increases as airspeed increases)
  3. Induced drag remains constant at all airspeeds between stall and Vne
  4. Induced drag increases linearly with True Airspeed

Explanation

Induced drag is a byproduct of lift generation: Di ∝ 1 / V². At low airspeeds, the aircraft must fly at a high angle of attack, creating strong wingtip vortices and high downwash, resulting in high induced drag. As airspeed increases in level flight, required angle of attack decreases, so induced drag drops rapidly.

Question 39Performance, loading and principles of flight

What aerodynamic phenomenon causes adverse yaw when a pilot deflects the ailerons to initiate a turn?

  1. The down-going aileron increases wing camber and lift, generating higher induced drag on the rising wing, which yaws the aircraft nose away from the direction of bank Correct answer.
  2. The up-going aileron increases lift and induced drag on the lowering wing, yawing the nose into the turn
  3. Both ailerons increase lift equally, but the outer wing travels faster and yaws the aircraft away from the turn
  4. The down-going aileron reduces camber and drag on the rising wing, yawing the nose away from the turn

Explanation

To bank right, the left aileron deflects downward, increasing camber and angle of attack on the left wing. This produces more lift but simultaneously creates significantly higher induced drag on the left (rising) wing, yawing the nose to the left (opposite to the intended turn). Rudder must be applied to coordinate the turn.

Question 40Performance, loading and principles of flight

What is the primary condition required for an aeroplane to enter an aerodynamic spin?

  1. Both wings must be stalled, with one wing stalled more deeply than the other, accompanied by uncoordinated yaw Correct answer.
  2. The aeroplane must be in a steep, unstalled spiral with both wings producing equal lift
  3. Both wings must stall symmetrically with no yaw or roll present
  4. One wing must be stalled while the other remains completely unstalled throughout the autorotation

Explanation

A spin is an autorotation condition that occurs when the aircraft is stalled (both wings beyond the critical angle of attack) and asymmetric yaw is present. One wing is stalled more deeply than the other, producing asymmetric lift and drag that sustains continuous autorotation in yaw and roll.

Question 41Performance, loading and principles of flight

In aircraft longitudinal stability, what is the 'Neutral Point', and how does its location relative to the Center of Gravity determine static stability?

  1. The aerodynamic center of the complete aeroplane; the CG must be positioned forward of the neutral point for the aeroplane to possess positive static longitudinal stability Correct answer.
  2. The aerodynamic center of the wing alone; the CG must coincide with it for positive static stability
  3. The aft CG limit at which stick force per g becomes zero; the CG must be aft of it for positive static stability
  4. The aerodynamic center of the complete aeroplane; placing the CG at that point provides positive static stability

Explanation

The Neutral Point is the aerodynamic center of the entire aircraft (wing, fuselage, and tail combined). If the center of gravity is forward of the neutral point, the static margin is positive and the aircraft possesses positive longitudinal static stability (a pitch disturbance generates a restoring moment).

Question 42Meteorology and weather information

Which combination most strongly favours the formation of radiation fog overnight?

  1. Moist air, a clear sky, light wind and strong nocturnal cooling over land Correct answer.
  2. Dry air, a thick cloud layer, strong wind and little surface cooling
  3. Warm moist air moving rapidly across a much colder sea surface
  4. A strong daytime sea breeze over a hot, dry inland surface

Explanation

Radiation fog develops when nocturnal terrestrial radiation cools moist air near the ground to its dew point. Clear skies maximise radiative cooling and a light wind mixes the cooling through a shallow layer without dispersing it.

Question 43Meteorology and weather information

What weather is normally associated with a mature occlusion in a mid-latitude depression?

  1. Extensive layered cloud and precipitation near the occluded front, with the warm sector lifted from the surface Correct answer.
  2. A narrow line of vigorous showers followed immediately by a broad surface warm sector
  3. A mainly cloud-free ridge with subsiding air and good visibility at every level
  4. Warm-front cloud ahead but no precipitation or cloud near the point of occlusion

Explanation

An occlusion forms when the faster-moving cold front catches the warm front and lifts the intervening warm air from the surface. The resulting frontal zone commonly produces widespread stratiform cloud and precipitation, with embedded convective cloud possible depending on stability.

Question 44Meteorology and weather information

At the standard mid-latitude tropopause, how do temperature lapse rate and typical jet-stream position change compared with the troposphere below?

  1. The temperature lapse rate becomes very small or reverses, and strong jet-stream winds are commonly found near the tropopause Correct answer.
  2. Temperature continues decreasing at the normal tropospheric lapse rate, while the strongest winds remain near the surface
  3. A strong temperature inversion begins at every tropopause, while jet streams are normally centred far below it
  4. Temperature becomes isothermal, but jet-stream winds are normally confined to the middle stratosphere

Explanation

The tropospheric decrease of temperature with height largely stops at the tropopause, which is an approximately isothermal transition to the stratosphere. The strongest mid-latitude jet streams are commonly concentrated near this boundary where horizontal temperature gradients are large.

Question 45Meteorology and weather information

What are the standard mean sea level (MSL) atmospheric values defined by the ICAO Standard Atmosphere (ISA)?

  1. Pressure: 1,013.25 hPa; Temperature: +15°C; Density: 1.225 kg/m³; Temperature lapse rate: 2°C (1.98°C) per 1,000 ft up to 11 km (36,090 ft) Correct answer.
  2. Pressure: 1,000.00 hPa; Temperature: +20°C; Density: 1.000 kg/m³; Temperature lapse rate: 3°C per 1,000 ft up to 10 km
  3. Pressure: 1,013.25 hPa; Temperature: 0°C; Density: 1.293 kg/m³; Temperature lapse rate: 1.5°C per 1,000 ft up to 8 km
  4. Pressure: 1,020.00 hPa; Temperature: +15°C; Density: 1.150 kg/m³; Temperature lapse rate: 0.65°C per 100 ft up to 15 km

Explanation

ICAO Standard Atmosphere (ISA) defines MSL conditions as: Pressure = 1,013.25 hPa (29.92 inHg), Temperature = +15°C (288.15 K), Density = 1.225 kg/m³, with a standard tropospheric temperature lapse rate of 1.98°C (commonly rounded to 2°C) or 0.65°C/100 m up to 11 km (36,090 ft), above which temperature is constant at -56.5°C.

Question 46Meteorology and weather information

An aerodrome has an elevation of 750 ft AMSL. The local QNH is 1001 hPa. What is the approximate pressure altitude of the aerodrome?

  1. 1,110 ft Correct answer.
  2. 750 ft
  3. 390 ft
  4. 1,350 ft

Explanation

Standard datum pressure is 1,013 hPa. The difference from QNH is 1,013 - 1,001 = 12 hPa. Using the standard 30 ft/hPa lapse rate, the datum is 12 * 30 = 360 ft below sea level. Pressure altitude = Elevation + (1,013 - QNH) * 30 = 750 + 360 = 1,110 ft.

Question 47Meteorology and weather information

What is the primary physical mechanism that initiates the dissipation stage of a single-cell (airmass) thunderstorm?

  1. The downdraught created by falling precipitation spreads throughout the entire cloud cell, cutting off the supply of warm, moist updraught air Correct answer.
  2. The updraught first reaches the tropopause and spreads into an anvil, immediately ending all vertical motion
  3. Latent heat released by freezing strengthens the updraught until precipitation can no longer fall
  4. Evaporation below cloud base warms the downdraught until it reverses into an updraught

Explanation

In an airmass thunderstorm, the mature stage produces heavy precipitation that drags cold air downward. In the dissipation stage, this cold downdraught expands across the entire base of the storm, cutting off the warm, moist updraught inflow required to feed the convective engine, causing the storm to collapse.

Question 48Meteorology and weather information

A moist air parcel with a temperature of +18°C at sea level is forced to rise over a mountain ridge 2,000 m high. Cloud base is reached at 1,000 m. The dry adiabatic lapse rate (DALR) is 1°C/100 m and the saturated adiabatic lapse rate (SALR) is 0.5°C/100 m. All condensed water falls out as rain. What is the temperature of the air when it descends back to sea level on the leeward side (Foehn effect)?

  1. +23°C Correct answer.
  2. +18°C
  3. +13°C
  4. +28°C

Explanation

Ascent: Sea level to 1,000 m is dry: 1,000 m * 1°C/100 m = 10°C drop → Temp at 1,000 m = 18 - 10 = +8°C. From 1,000 m to 2,000 m is saturated: 1,000 m * 0.5°C/100 m = 5°C drop → Temp at crest (2,000 m) = 8 - 5 = +3°C. Descent: As all moisture was precipitated out, descent is dry all the way from 2,000 m to sea level: 2,000 m * 1°C/100 m = +20°C rise. Sea level leeward temp = +3 + 20 = +23°C.

Question 49Meteorology and weather information

Decode the following METAR extract: 'EKCH 051120Z 28018G32KT 240V310 1200 R04L/1400D +SHSN FEW008 BKN015CB 01/M01 Q0998='

  1. Wind 280° true at 18 kt gusting 32 kt varying between 240° and 310°; visibility 1,200 m; RVR runway 04L 1,400 m downward trend; heavy snow showers; few clouds at 800 ft, broken cumulonimbus at 1,500 ft; temperature +1°C, dew point -1°C; QNH 998 hPa Correct answer.
  2. Wind 280° magnetic at 18 kt gusting 32 kt; visibility 12 km; runway 04L dry; moderate snow showers; few clouds at 8,000 ft; QNH 998 hPa
  3. Wind 240° to 310° at 18 kt; visibility 1,200 ft; RVR 1,400 ft decreasing; light snow; overcast cumulonimbus at 1,500 ft; QNH 998 hPa
  4. Wind 280° true at 32 kt; visibility 1,200 m; RVR runway 04L 1,400 m with no change; thunderstorm with snow; temperature -1°C; QNH 1098 hPa

Explanation

Correct decoding: 28018G32KT = wind 280° true, 18 kt gusts 32 kt; 240V310 = direction varying between 240° and 310°; 1200 = prevailing visibility 1,200 m; R04L/1400D = Runway 04L RVR is 1,400 m with downward tendency (D); +SHSN = heavy snow showers; FEW008 BKN015CB = few clouds at 800 ft AAL, broken cumulonimbus at 1,500 ft AAL; 01/M01 = temp +1°C, dew point -1°C; Q0998 = QNH 998 hPa.

Question 50Meteorology and weather information

What is the meteorological definition of a low-level microburst associated with convective activity, and what flight hazard does it pose during landing approach?

  1. A concentrated, violent downdraught of less than 4 km horizontal extent producing intense outflow winds; an aircraft encounters an initial headwind increasing airspeed, followed immediately by a severe downdraught and strong tailwind, causing rapid loss of altitude and airspeed Correct answer.
  2. A broad convective downdraught more than 20 km across whose wind changes are gradual over several minutes
  3. A gust front with only a horizontal wind-direction change and no associated vertical velocity
  4. A small dust devil whose circulation remains below the aircraft throughout the approach

Explanation

A microburst is an intense convective downdraught covering a horizontal diameter of less than 4 km (2.2 NM) with duration typically under 15 minutes. Upon entering, an aeroplane first encounters an increasing headwind (performance gain), which tempts the pilot to reduce power. Seconds later, the aircraft enters the strong downdraught followed by a severe tailwind (performance loss), driving the aircraft into the ground.