Correct: C)
Explanation: The correct answer is C because to convert pounds to kilograms, divide by 2.2: 1235 / 2.2 = 561.4 kg, which rounds to approximately 560 kg. A (620 kg) would correspond to about 1364 lbs. B (2720 kg) results from multiplying instead of dividing. D (2470 kg) is also the result of a multiplication error. The key formula is: mass in kg = weight in lbs / 2.2.
Correct: C)
Explanation: The correct answer is C because on an upsloping field with a tailwind, the competing effects partially cancel each other: the upslope shortens the ground roll while the tailwind lengthens it. The normal approach speed (yellow triangle on the ASI) provides the correct balance of energy management. A is wrong because a faster approach would result in excessive float on the upslope. B is wrong because flaring higher risks ballooning on the slope. D is wrong because full airbrakes may cause an excessively steep descent on short final.
Correct: A)
Explanation: The correct answer is A because at 2000 m AMSL above Langenthal, you are in Class E airspace. VFR flight in Class E requires 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance. B is wrong because Class G with its reduced minima applies only at very low altitudes. C is wrong because there is no Class D TMA at this location and altitude. D is wrong because Class C begins at FL 130 in this region, far above 2000 m AMSL.
Correct: C)
Explanation: The correct answer is C because when the C.G. is too far aft, the glider loses longitudinal static stability — the nose tends to pitch up without returning to equilibrium, potentially leading to uncontrollable divergent oscillations or a stall/spin. A (too far forward) is less dangerous because the aircraft remains stable, though elevator authority may be insufficient for landing. B and D are wrong because vertical C.G. displacement is not the primary concern in standard glider mass-and-balance analysis.
Correct: C)
Explanation: The correct answer is C because the airspeed indicator measures dynamic pressure, which inherently accounts for air density. The V_NE marking on the ASI (red line) represents a fixed IAS value that corresponds to the structural limit. However, note that the allowable maximum IAS must actually be reduced at high altitude per the flight manual's speed-altitude table — the ASI marking itself does not change, but the pilot must observe a lower limit. A and B/D are wrong because the physical mark on the instrument does not move. The subtlety is that while the ASI reading mechanism inherently accounts for density, glider pilots must consult the altitude-correction table for the actual limit at high altitude.
Correct: C)
Explanation: The correct answer is C because ground speed = distance / time = 150 km / 1.25 hours = 120 km/h. The key step is converting 1 hour 15 minutes to decimal hours: 15 minutes = 0.25 hours, so total time = 1.25 hours. A (125 km/h) results from dividing by 1.2 hours. B (115 km/h) and D (110 km/h) do not correspond to any correct calculation with these inputs.
]
- A) The extended CTR/TMA Payerne and restricted zone LS-R4 must be strictly avoided every day from 02 to 06 September 2013, between sunrise and sunset.
- B) An airshow is taking place in the Payerne area from 02 to 06 September 2013. The TMA Payerne and restricted zone LS-R4 are active each day during this period between 0600 UTC and 1500 UTC as holding areas and airshow demonstration sectors.
- C) Due to an airshow from 02 to 06 September 2013, the extended CTR/TMA Payerne is active each day between 0600 UTC and 1500 UTC. The TMA is used as a holding area, the restricted zone LS-R4 as a demonstration and holding area. The area must be strictly avoided.
- D) Due to an airshow, a transit clearance for the extended CTR/TMA Payerne and restricted zone LS-R4 must be requested on frequency 135.475 (Payerne TWR) from 02 to 06 September 2013.
Correct: C)
Explanation: The correct answer is C because the NOTAM establishes that from 2 to 6 September 2013, between 0600 and 1500 UTC, the extended CTR/TMA Payerne is activated as a holding area, while LS-R4 serves as both a demonstration and holding area for an airshow. These areas must be strictly avoided during the active period. A is wrong because the times are 0600-1500 UTC, not sunrise to sunset. B incorrectly states both areas serve as holding and demonstration areas. D is wrong because transit is not permitted — the area must be avoided entirely, not transited with clearance.
]
- A) 95km/h
- B) 75km/h
- C) 55km/h
- D) 135km/h
Correct: B)
Explanation: The correct answer is B (75 km/h) because the best glide speed is found by drawing a tangent from the origin to the speed polar curve for 450 kg. The point where this tangent touches the curve gives the speed for maximum lift-to-drag ratio (best glide). A (95 km/h) is too fast and would correspond to a heavier mass or a different polar. C (55 km/h) is near the stall speed. D (135 km/h) is deep in the high-speed range where the glide ratio is significantly reduced.
]
- A) The DABS can be ignored as it solely applies to military aircraft.
- B) You may pass through all relevant danger and restricted zones below 1000 ft AGL or above 10,000 ft AMSL.
- C) The route can be flown without coordination between 1200 and 1300 LT.
- D) It is not possible to fly the planned route that day.
Correct: C)
Explanation: The correct answer is C because checking the DABS for 19 March 2013 (winter time, CET = UTC+1), the planned time of 1205-1255 LT converts to 1105-1155 UTC. During this period, the relevant danger and restricted zones along the route are not active, allowing the route to be flown without coordination. A is wrong because the DABS applies to all airspace users, including gliders. B is wrong because altitude-based exemptions do not automatically apply to all restricted areas. D is wrong because the route is flyable during the specified time window.
Correct: A)
Explanation: The correct answer is A because stall speed (and therefore minimum speed) is proportional to the square root of wing loading. If wing loading increases by 40% (factor 1.4), the new minimum speed is the original multiplied by the square root of 1.4, which equals approximately 1.183 — an increase of about 18.3%. B is wrong because the speed does not increase linearly with wing loading. C is wrong because a 100% increase would mean doubling the speed. D is wrong because any mass increase raises the minimum speed.
]
- A) the sink rate of the ASK21 is independent of its mass
- B) the ASK21 has a worse glide ratio at lower flying mass
- C) the ASK21 has a higher sink rate at higher flying mass
- D) the ASK21 has a better glide ratio at lower flying mass
Correct: A)
Explanation: The correct answer is A because at 150 km/h, the two polar curves for different masses of the ASK21 intersect, meaning both configurations have the same sink rate at this particular speed. This is an aerodynamic property of the polar: the curves cross at one speed where mass has no effect on sink rate. B is wrong because at 150 km/h the glide ratio is equal for both masses. C is wrong because the sink rates are identical at this intersection point. D is also wrong because neither mass has a better glide ratio at this specific speed.
]
- A) 700 ft.
- B) 780m.
- C) 780 ft
- D) 700m.
Correct: B)
Explanation: The correct answer is B (780 m) because the AIP chart for Amlikon aerodrome shows a maximum landing distance available of 780 metres in the eastward direction. A and C are wrong because landing distances in Switzerland are given in metres, not feet. D (700 m) does not match the published data for the eastward heading. Always verify the unit and the specific runway direction when reading aerodrome charts.
]
- A) 2400 ft AMSL.
- B) 3500 ft AMSL.
- C) 2000ft GND.
- D) 5000 ft AMSL.
Correct: B)
Explanation: The correct answer is B because the EMMEN TMA lower boundary between Cham and Hitzkirch is at 3500 ft AMSL. Below this altitude, you remain in uncontrolled airspace and no clearance is needed. Above 3500 ft AMSL, you enter the TMA and must obtain an ATC clearance. A (2400 ft) is too low and does not correspond to the published limit. C (2000 ft GND) references above ground level, which is not how this TMA boundary is expressed. D (5000 ft) is too high.
Correct: D)
Explanation: The correct answer is D because when the maximum permitted payload is exceeded, the only correct action is to reduce the payload until it complies with the limit. The maximum payload is a certification limit based on structural strength and C.G. envelope. A and C are wrong because trimming adjusts aerodynamic forces on the tail but does not change the aircraft's mass or C.G. — it cannot make an overloaded aircraft safe. B is wrong because increasing takeoff speed does not solve an overweight condition and may actually overstress the structure further.
Correct: D)
Explanation: The correct answer is D because a headwind reduces groundspeed while the sink rate in the airmass remains unchanged. Since the glider covers less horizontal ground distance per unit of altitude lost, the descent angle relative to the ground steepens (increases). A is wrong because a tailwind decreases (flattens) the glide angle over the ground by increasing groundspeed. B is wrong because a headwind increases, not decreases, the ground glide angle. C is wrong because wind significantly affects the ground track glide angle, even though it does not affect the airmass glide angle.
Correct: B)
Explanation: The correct answer is B because as altitude increases, air density decreases. For the same true airspeed, the pitot tube measures less dynamic pressure, so the IAS reading is lower than TAS. Conversely, to maintain the same IAS at altitude, the aircraft must fly at a higher TAS. The relationship is approximately TAS = IAS x square root of (sea-level density / actual density). A is wrong because IAS does not rise relative to TAS with altitude. C is wrong because IAS can always be measured. D is wrong because IAS and TAS diverge increasingly with altitude.
Correct: A)
Explanation: The correct answer is A because heavy rain on the wing surface increases roughness and can degrade the boundary layer, potentially raising the stall speed and reducing maximum lift coefficient. A higher approach speed provides a safety margin against these effects. B is wrong because deliberately increasing wing loading in rain would require adding ballast, which is impractical and counterproductive. C is wrong because a shallower approach reduces obstacle clearance in poor visibility. D is wrong because a lower approach speed reduces the safety margin when aerodynamic degradation is already a risk.
]
- A) The traffic pattern for runway 33 is clockwise.
- B) The traffic pattern for runway 15 is clockwise.
- C) The traffic pattern for runway 33 is counter-clockwise.
- D) Depending on wind, the traffic pattern for runway 33 may be either clockwise or counter-clockwise.
Correct: D)
Explanation: The correct answer is D because at Bex aerodrome, terrain constraints (the Rhone valley and surrounding mountains) mean the traffic pattern direction for runway 33 depends on the prevailing wind conditions. The chart shows that either a left or right circuit may be used. A is wrong because it limits the pattern to clockwise only. B relates to runway 15, not 33. C is wrong because it limits the pattern to counter-clockwise only. Pilots must check the local procedures and wind conditions before joining the circuit.
]
- A) 3500 ft AGL.
- B) FL 100.
- C) FL 35.
- D) 3500 ft AMSL.
Correct: D)
Explanation: The correct answer is D because the lower limit of TMA BERN 1 over Biel Kappelen is 3500 ft AMSL. By staying below this altitude, you remain in uncontrolled airspace and do not need a transit clearance. A (3500 ft AGL) is wrong because TMA boundaries are referenced to MSL, not AGL. B (FL 100) is far above the relevant boundary. C (FL 35) converts to approximately 3500 ft in standard atmosphere, but flight levels use the standard pressure setting (1013.25 hPa), not QNH, so this is not the correct way to express the limit.
Correct: A)
Explanation: The correct answer is A because applying the mass-and-balance calculation with the data provided (from the attached sheet), the new C.G. position computes to 76.7, which falls within the approved forward and aft C.G. limits. B (78.5) is an incorrect calculation result. C (82.0) is too far aft and would be outside limits. D (75.5) is incorrectly calculated and would also fall outside the forward limit. Always verify your calculation by checking whether the result is between the published forward and aft limits.
Correct: A)
Explanation: The correct answer is A because a waterlogged grass surface creates greater friction and drag on the landing gear during the ground roll, causing the glider to decelerate faster and stop in a shorter distance. The water acts as a braking medium. B is wrong because wet grass increases, not decreases, rolling resistance for a glider. C is wrong because while directional control may be slightly affected, the primary effect is shortened stopping distance. D is wrong because surface conditions always affect landing distance.
]
- A) 520 m.
- B) 470m.
- C) 520 ft.
- D) 470 ft.
Correct: B)
Explanation: The correct answer is B (470 m) because the AIP chart for Schanis aerodrome shows a maximum landing distance available of 470 metres in the NNW direction. A (520 m) does not match the published data for this heading. C and D are wrong because Swiss aerodrome distances are given in metres, not feet. Always read the correct runway direction and corresponding distance from the aerodrome chart.
Correct: D)
Explanation: The correct answer is D (45.71 lbs). The calculation uses the shift formula: when mass x is moved from the current C.G. position (80) to arm 150, the C.G. shifts aft. The new C.G. must not exceed 80.5. Using the formula: delta CG = (x × delta arm) / total mass, we get: 0.5 = (x × 70) / 6400, therefore x = (0.5 × 6400) / 70 = 45.71 lbs. A (27.82), B (56.63), and C (39.45) result from incorrect calculations using wrong distances or mass values.
Correct: C)
Explanation: The correct answer is C because correct loading requires satisfying two independent conditions simultaneously: the total mass must not exceed the maximum allowable mass (MTOM), and the payload must be distributed so that the C.G. remains within the approved envelope. A is wrong because respecting the mass limit alone does not guarantee the C.G. is within limits. B is wrong because correct distribution alone does not ensure the total mass is within limits. D is wrong because it addresses only one specific baggage compartment rather than the complete loading requirements.
]
- A) in the speed range up to 100 km/h, an increase in flying mass reduces the sink rate.
- B) minimum speed is independent of flying mass.
- C) both glide ratio and minimum speed are independent of flying mass.
- D) only the maximum glide ratio is independent of flying mass, apart from a minor Reynolds number effect.
Correct: D)
Explanation: The correct answer is D because when comparing polar curves for different masses, the tangent from the origin touches each curve at the same angle, meaning the maximum lift-to-drag ratio (best glide ratio) is essentially unchanged by mass, apart from minor Reynolds number effects. However, the speed at which this best glide ratio occurs increases with mass. A is wrong because increasing mass always increases the sink rate at any given speed. B is wrong because minimum speed increases with mass (proportional to the square root of mass ratio). C is wrong because while glide ratio is mass-independent, minimum speed is not.
Correct: A)
Explanation: The correct answer is A because the airspeed indicator measures dynamic pressure, which directly relates to aerodynamic forces regardless of altitude. At 1800 m AMSL, air density is lower, so the TAS will be higher for the same IAS — but the aerodynamic forces (lift, stall characteristics) depend on IAS, not TAS. Therefore, the same indicated approach speed provides the same safety margins as at sea level. B is wrong because flying at a lower IAS would reduce the stall margin. D is wrong because a higher IAS is unnecessary and would result in excessive float. C is wrong because the minimum sink speed is not the correct approach speed.
]
- A) 130km/h
- B) 90km/h
- C) 70km/h
- D) 110km/h
Correct: B)
Explanation: The correct answer is B (90 km/h) because the best glide ratio speed is found where the tangent from the origin touches the speed polar curve for 450 kg. For this glider type at 450 kg, this occurs at approximately 90 km/h. A (130 km/h) is too fast — at this speed the glide ratio is significantly reduced. C (70 km/h) is closer to the minimum sink speed, which maximises endurance but not distance. D (110 km/h) would give a reduced glide ratio compared to the optimum.
Correct: C)
Explanation: The correct answer is C because when the aft C.G. limit is exceeded, the useful load must be redistributed to move mass forward — for example, adding nose ballast, repositioning equipment, or adjusting the pilot's seating position. This physically moves the C.G. within approved limits. A is wrong because trimming aft would worsen the situation aerodynamically. B is wrong because being within mass limits does not compensate for a C.G. out of limits — both must be satisfied independently. D is wrong because trim adjusts aerodynamic forces but does not change the actual C.G. position.
Correct: D)
Explanation: The correct answer is D because high temperature reduces air density, decreasing the lift generated at any given groundspeed, requiring a longer acceleration to reach flying speed. A tailwind reduces the headwind component, meaning the aircraft needs a higher groundspeed to achieve the same airspeed, further lengthening the takeoff run. A is wrong because low temperature increases air density (more lift) and headwind shortens the run. B is wrong because a strong headwind shortens the takeoff distance. C is wrong because high atmospheric pressure increases density, which helps rather than hinders takeoff performance.
]
- A) On 18 November, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. Lower limit: Class E airspace, upper limit: max. FL150.
- B) On 18 November from 1800 LT to 2100 LT, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas.
- C) On 18 November from 1800 UTC to 2100 UTC, a military night flying exercise with helicopters will take place.
- D) On 18 November from 1800 UTC to 2100 UTC, a military night flying exercise will take place in the ZUGERSEE, SUSTEN and TICINO areas. Lower limit: GND, upper limit: max. 15,000 ft AMSL.
Correct: D)
Explanation: The correct answer is D because the NOTAM specifies a military night flying exercise on 18 November from 1800 to 2100 UTC in the ZUGERSEE, SUSTEN, and TICINO areas, with vertical limits from GND to 15,000 ft AMSL. A is wrong because the lower limit is GND, not Class E airspace, and the upper limit is 15,000 ft AMSL, not FL150. B is wrong because the times are in UTC, not local time. C is wrong because it incorrectly specifies helicopter-only operations and omits the geographic areas.
]
- A) 5500 ft GND.
- B) 4500 ft AMSL.
- C) 5000 ft AMSL
- D) 3000 ft AMSL.
Correct: D)
Explanation: The correct answer is D because the CTR (Control Zone) of Bern-Belp airport has an upper limit of 3000 ft AMSL. Above this altitude, you exit the CTR and enter different airspace. A (5500 ft GND) does not match the published limit. B (4500 ft AMSL) is too high. C (5000 ft AMSL) is also too high. VFR flight within the CTR requires a clearance from Bern Tower and must remain below the published upper limit.
]
- A) Class G airspace, horizontal visibility 1.5 km, clear of cloud with continuous sight of the ground.
- B) Class C airspace, horizontal visibility 8 km, cloud clearance 1.5 km horizontally, 300 m vertically.
- C) Class C airspace, horizontal visibility 5 km, cloud clearance 1.5 km horizontally, 300 m vertically.
- D) Class E airspace, horizontal visibility 5 km, cloud clearance 1.5 km horizontally, 300 m vertically.
Correct: D)
Explanation: The correct answer is D because at 1700 m AMSL above Bex aerodrome, you are in Class E airspace. VFR minima in Class E require 5 km horizontal visibility, 1500 m horizontal cloud clearance, and 300 m vertical cloud clearance. A is wrong because Class G applies at lower altitudes with reduced requirements. B is wrong because Class C has the right visibility minimum (5 km in Switzerland, not 8 km) but starts at a much higher altitude. C is wrong for the same airspace classification reason — Class C begins at FL 130, well above 1700 m.
]
- A) 1,6m/s
- B) 0,8m/s
- C) 2,0m/s
- D) 1,2m/s
Correct: C)
Explanation: The correct answer is C (2.0 m/s) because reading the speed polar curve for a flying mass of 580 kg at 160 km/h, the sink rate is approximately 2.0 m/s. A (1.6 m/s) would correspond to a lighter mass or lower speed. B (0.8 m/s) is near the minimum sink rate at much lower speed. D (1.2 m/s) is also too low for this speed and mass combination. When reading a speed polar, always identify the correct curve for the given mass before reading the value at the specified speed.
Correct: B)
Explanation: The correct answer is B because to convert kilograms to pounds, multiply by 2.2: 550 x 2.2 = 1,210 lbs. A (12,100 lbs) results from multiplying by 22 instead of 2.2. C (2,500 lbs) does not correspond to any correct calculation. D (250 lbs) results from dividing instead of multiplying. The key formula is: weight in lbs = mass in kg x 2.2.
Correct: D)
Explanation: The correct answer is D because the best glide ratio speed (also called best L/D speed) maximises the horizontal distance covered per unit of altitude lost in still air. This speed is found on the polar curve where the tangent from the origin touches the curve. A is wrong because minimum sink speed maximises endurance (time aloft), not distance. B is wrong because maximum speed produces the worst glide ratio due to high parasite drag. C is wrong because minimum flying speed is near the stall and gives a poor glide ratio due to high induced drag.
Correct: A)
Explanation: The correct answer is A because the maximum glide ratio (best L/D) is essentially independent of mass — both the lift coefficient and drag coefficient at the optimal angle of attack remain the same, so their ratio is unchanged. Only a minor Reynolds number effect exists. B is wrong because wing loading = mass / wing area, which directly increases with mass. C is wrong because sink rate increases with mass at any given speed. D is wrong because the speeds corresponding to best glide and minimum sink both increase with mass.
Correct: D)
Explanation: The correct answer is D because time = distance / speed = 150 km / 100 km/h = 1.5 hours = 1 hour 30 minutes. A (1 hour 50 minutes) would correspond to a distance of about 183 km. B (1 hour 40 minutes = 1.667 hours) would correspond to about 167 km. C (2 hours) would correspond to 200 km. The calculation is straightforward: 150 / 100 = 1.5 hours. Convert the decimal 0.5 hours to 30 minutes.
]
- A) The route can be flown without restriction after contacting 128.375 MHz.
- B) Restricted zones LS-R8 and LS-R8A may be transited below 28,000 ft AMSL.
- C) It is not possible to fly this route while the restricted zones are active.
- D) Restricted zones LS-R8 and LS-R8A may be overflown at 9200 ft AMSL or above.
Correct: C)
Explanation: The correct answer is C because when restricted zones LS-R8 and LS-R8A are active, they cover the planned alpine route between Munster and Amsteg, making it impossible to fly through them. Restricted zones with "entry prohibited" status cannot be transited, regardless of altitude or radio contact. A is wrong because radio contact does not grant transit rights through active restricted zones. B is wrong because a 28,000 ft ceiling does not help a glider. D is wrong because overflying at 9,200 ft may still be within the zone's vertical limits.
Correct: A)
Explanation: The correct answer is A because to transit the Zurich TMA, the pilot must make first radio contact on frequency 124.7 MHz (Zurich Information) at least 10 minutes before entering the controlled airspace. This provides ATC sufficient time to assess traffic, issue a clearance or alternative instructions, and ensure separation. B is wrong because 5 minutes is insufficient lead time. C is wrong because 118.975 is not the correct frequency for Zurich TMA transit requests. D is wrong on both the frequency and the lead time.
Correct: A)
Explanation: The correct answer is A because in a turn, the stall speed increases by the square root of the load factor: Vsturn = Vsstraight x sqrt(n). With n = 2.0: Vs_turn = 60 x sqrt(2) = 60 x 1.414 = 84.85 kts. The increase is (84.85 - 60) / 60 x 100 = 41.4%, which rounds to approximately 40%. B is wrong because the stall speed always increases in a turn. C (5%) and D (20%) significantly underestimate the effect. This relationship between bank angle, load factor, and stall speed is fundamental to safe manoeuvring flight.
Correct: C)
Explanation: The correct answer is C because restricted airspace areas (LO R) on aeronautical charts express their limits using standard altitude references. LO R 16 has an upper limit of 1,500 ft MSL (mean sea level), which is a fixed, absolute altitude. A is wrong because 1,500 m MSL would be approximately 4,900 ft — a completely different altitude that confuses feet with metres. B is wrong because FL150 (15,000 ft pressure altitude) is far too high for a typical low-level restriction. D is wrong because 1,500 ft GND (above ground level) would vary with terrain elevation and is not the published reference.
Correct: B)
Explanation: The correct answer is B because LO R 4 has its upper limit at 4,500 ft MSL, a fixed altitude above mean sea level. A is wrong because 4,500 ft AGL (above ground level) would vary with terrain, which is inappropriate for a fixed regulatory boundary. C is wrong because 1,500 ft AGL is both the wrong altitude value and the wrong reference. D is wrong because 1,500 ft MSL is too low and corresponds to a different restricted area (LO R 16).
Correct: B)
Explanation: The correct answer is B because the NOTAM prohibits overflight up to an altitude of 9,500 ft MSL, following ICAO convention where "altitude" refers to height above mean sea level. A is wrong because "height" in aviation terminology means above a local ground reference (AGL), which is not what the NOTAM specifies. C is wrong because FL 95 is a pressure altitude reference based on 1013.25 hPa, which differs from an MSL altitude depending on actual atmospheric conditions. D is wrong because 9,500 m MSL would be approximately 31,000 ft — clearly inconsistent with a typical VFR NOTAM.
Correct: B)
Explanation: The correct answer is B (symbol C in the annex) because ICAO aeronautical chart symbology (defined in ICAO Annex 4) uses specific symbols to distinguish between single and grouped obstacles, and between lighted and unlighted ones. Symbol C represents a group of unlighted obstacles. A (symbol D), C (symbol B), and D (symbol A) represent other obstacle categories such as single obstacles, lighted groups, or lighted single obstacles. Correct identification of these symbols is essential for cross-country flight planning and obstacle avoidance.
Correct: B)
Explanation: The correct answer is B (symbol A in the annex) because ICAO chart symbology uses distinct depictions for different aerodrome types — civil versus military, international versus domestic, and paved versus unpaved. Symbol A represents a civil (non-international) airport with a paved runway. A (symbol D), C (symbol C), and D (symbol B) represent other aerodrome categories such as international airports, military aerodromes, or grass-strip airfields. Glider pilots must recognise these symbols when identifying potential emergency landing options.
Correct: D)
Explanation: The correct answer is D (symbol C in the annex) because on ICAO aeronautical charts, a general spot elevation is indicated by a specific symbol showing a terrain point of known height, used for situational awareness and terrain clearance planning. A (symbol A), B (symbol B), and C (symbol D) represent other elevation-related markings such as maximum elevation figures, surveyed points, or obstruction elevations defined in ICAO Annex 4.
Correct: A)
Explanation: The correct answer is A. The center of gravity is the single point through which the resultant of all gravitational forces acts on the aircraft — it is the mass-weighted average position of all components. B is wrong because the neutral point is a distinct aerodynamic concept used for stability analysis, not another name for C.G. C duplicates the same incorrect description as A's wording, but the C.G. is defined by mass distribution, not as a geometric midpoint. D is wrong because the C.G. is not the heaviest point — it is where the total weight effectively acts.
Correct: B)
Explanation: The correct answer is B because in mass-and-balance calculations, moment is defined as the product of mass and balance arm: Moment = Mass x Arm (e.g., in kg-m or lb-in). This follows the physical definition of a torque. The total C.G. is found by summing all moments and dividing by total mass. A is wrong because adding mass and arm is dimensionally meaningless. C is wrong because dividing mass by arm does not produce a moment. D is wrong because subtracting them is equally incorrect.
Correct: C)
Explanation: The correct answer is C because the balance arm (moment arm) is the horizontal distance measured from the aircraft's datum reference point to the center of gravity of a specific mass item. A is wrong because that describes the datum itself, not the balance arm. B is wrong because balance arms are measured from the datum, not from the overall aircraft C.G. D is wrong because that is the definition of the center of gravity of a mass item, not the balance arm.
Correct: D)
Explanation: The correct answer is D because interception lines (also called catching lines or line features) are prominent linear ground features — motorways, rivers, coastlines, railways — that a pilot selects during pre-flight planning to navigate toward if orientation is lost. By flying toward a known interception line, the pilot can re-establish position and resume navigation. A is wrong because interception lines are geographic features, not airport markers. B is wrong because they are not range indicators. C is wrong because nothing authorises continuing flight below VFR minima — interception lines are a lost-procedure tool, not a visibility workaround.
