Chapter 40

Revision Questions
DGCA CPL/ATPL Study Notes — Final Examination Preparation
Subject: Instrumentation | All Topics
Compiled by Capt. Pankaj Pahil
This chapter is a pure revision question bank covering the entire Instrumentation syllabus.
All questions are taken verbatim from the Oxford ATPL Ground Training Series source text.

Table of Contents

  1. Section A — Flight Instruments Questions (164 questions)
  2. Section B — Automatic Flight Questions (58 questions)
  3. Section C — Warning & Recording Questions (33 questions)
  4. Section D — Engine Instruments Questions (52 questions)
  5. Section E — Specimen Questions (55 questions with explanations)
  6. Section F — Specimen Examination Paper (55 questions)
  7. Master Answer Keys
How to use this chapter: This chapter contains the complete Oxford revision question bank for the Instrumentation subject. Each section contains the verbatim source questions with the source answer key. Selected key questions have detailed worked explanations. The Specimen Questions section (Section E) includes full explanations for all questions. Use the Master Answer Keys at the end to self-test.
Section A — Flight Instruments Questions
Coverage: Chapters 1–23 — Pitot-static systems, airspeed, altimetry, temperature, Machmeter, ADC, magnetism, compasses, gyroscopes, DGI, artificial horizon, turn indicators, remote magnetic compass, INS/IRS, radio altimeter, FMS, EFIS, and basic computers. 164 questions total.

Key Questions with Detailed Explanations

A-Q1.A 2-axis gyro measuring vertical changes will have:
  1. one degree of freedom, vertical axis
  2. two degrees of freedom, vertical axis
  3. one degree of freedom, horizontal axis
  4. two degrees of freedom, horizontal axis
Correct Answer: (b) two degrees of freedom, vertical axis
Explanation: A 2-axis gyro has two degrees of freedom (it can precess in two planes). To measure vertical changes (pitch and roll — earth vertical reference), the spin axis must be vertical. This is the configuration used in the artificial horizon. Degrees of freedom = number of axes around which the gyro can precess freely.
Why the other options are wrong:
  • (a) — One degree of freedom + vertical axis = rate gyro (measures rotation rate). A single-axis gyro cannot serve as a 2-axis instrument.
  • (c) — One DOF + horizontal axis = turn indicator gyro.
  • (d) — Two DOF + horizontal axis = DGI (Directional Gyro Indicator).
Instructor's Note: DOF = number of planes in which the gyro can precess. Artificial horizon: 2 DOF, vertical spin axis. DGI: 2 DOF, horizontal spin axis. Turn indicator: 1 DOF, horizontal spin axis. These pairings are fundamental gyroscope exam knowledge.
A-Q2.The properties of a gyro are: (1) mass (2) rigidity (3) inertia (4) precession (5) rotational speed
  1. 1, 2 & 3
  2. 2 & 4
  3. 2 & 3
  4. 1 & 3
Correct Answer: (b) 2 & 4 — Rigidity and Precession
Explanation: A gyroscope has two fundamental properties: (1) Rigidity in space — tendency to maintain its spin axis direction in space when spinning at high speed; and (2) Precession — when a force is applied, the gyro responds at 90° to the applied force in the direction of rotation. Mass, inertia, and rotational speed contribute to the degree of rigidity but are not the "properties" per se.
Why the other options are wrong:
  • (a), (c), (d) — Mass, inertia, and rotational speed affect gyroscopic performance but are not the two defining properties. Rigidity and precession are the two fundamental gyroscopic properties.
Instructor's Note: Gyro Properties = Rigidity + Precession. Everything else (mass, rpm, rotor diameter) affects the degree of rigidity. This two-word answer appears in virtually every DGCA instrumentation exam.
A-Q7.The Machmeter consists of:
  1. an airspeed indicator with Mach scale
  2. an airspeed indicator with an altimeter capsule
  3. an altimeter corrected for density
  4. a VSI and altimeter combined
Correct Answer: (b) an airspeed indicator with an altimeter capsule
Explanation: The Machmeter has two capsule systems: (1) a pitot-static capsule that measures dynamic pressure (like an ASI), and (2) an aneroid (altimeter) capsule that senses ambient (static) pressure as a proxy for altitude/density. The ratio of these two pressures gives Mach number: M = √(dynamic pressure / static pressure function). The altimeter capsule corrects the speed reading for altitude.
Why the other options are wrong:
  • (a) — A Mach scale alone does not constitute a Machmeter; the additional altitude-compensating capsule is the key differentiator.
  • (c) — An altimeter corrected for density gives density altitude, not Mach number.
  • (d) — VSI + altimeter gives rate of climb, not Mach number.
Instructor's Note: Mach = TAS ÷ LSS. Since LSS varies with temperature (and thus altitude), the Machmeter needs both an ASI capsule (dynamic pressure) and an altimeter capsule (static pressure). The mechanical linkage between these two capsules outputs Mach number directly.
A-Q10.An aircraft is flying at an indicated altitude of 16,000 ft. The outside air temperature is −30°C. What is the true altitude of the aircraft?
  1. 16,200 ft
  2. 15,200 ft
  3. 18,600 ft
  4. 13,500 ft
Correct Answer: (b) 15,200 ft
Explanation: ISA temperature at 16,000 ft = 15 − (16 × 2) = 15 − 32 = −17°C. Actual OAT = −30°C. Temperature deviation = −30 − (−17) = −13°C colder than ISA. True altitude = indicated altitude × (actual temp / ISA temp in Kelvin). Using the 4 ft/°C/1000 ft rule: correction = −13 × 16 × 4 / 1000 ≈ −832 ft. True alt ≈ 16,000 − 832 ≈ 15,168 ft ≈ 15,200 ft. Cold temperatures make the aircraft lower than indicated — the altimeter over-reads.
Why the other options are wrong:
  • (a) — 16,200 ft would require the aircraft to be in warmer than ISA air.
  • (c) — 18,600 ft is not achievable with these temperature conditions.
  • (d) — 13,500 ft overestimates the cold-air correction.
Instructor's Note: Cold air = denser air = altimeter over-reads = true altitude is LESS than indicated. Warm air = less dense = altimeter under-reads = true altitude is MORE than indicated. Memory: "Cold and Low, Warm and High" — pilots flying in cold air are lower than their altimeter shows.
A-Q12.QNH is:
  1. the airfield barometric pressure
  2. the setting that will give zero indication on the airfield
  3. the equivalent sea level pressure at the airfield
  4. the setting that will indicate airfield height
Correct Answer: (c) the equivalent sea level pressure at the airfield
Explanation: QNH is the altimeter sub-scale setting that causes the altimeter to read the airfield's altitude above mean sea level (AMSL) when the aircraft is on the ground. It is derived by reducing the airfield barometric pressure to MSL using the ISA lapse rate. Setting QNH gives altitude AMSL, not height above the airfield (that would be QFE). Option (b) describes QFE.
Why the other options are wrong:
  • (a) — The actual airfield barometric pressure is QFE (gives height above airfield).
  • (b) — Zero indication at the airfield = QFE setting (height above aerodrome).
  • (d) — QNH indicates altitude AMSL (e.g., 1,400 ft), not height above airfield (which would be zero at ground level if QFE were set).
Instructor's Note: QNH = altitude AMSL. QFE = height above aerodrome. QNE = altitude at standard pressure (1013.25 hPa), used as flight level. These three Q-codes and their definitions are fundamental DGCA exam topics.
A-Q13.What is the Schuler period?
  1. 21 minutes
  2. 84 minutes
  3. 1 oscillation in azimuth
  4. 63 minutes
Correct Answer: (b) 84 minutes
Explanation: The Schuler period is 84.4 minutes — the natural oscillation period of an inertial navigation platform when tuned to Earth's radius. It is derived from the pendulum equation: T = 2π√(R/g), where R = Earth's radius (~6,371 km) and g = gravitational acceleration. An INS tuned to the Schuler period will not precess due to accelerations caused by motion over the Earth's curved surface.
Why the other options are wrong:
  • (a) — 21 minutes is not a standard INS parameter.
  • (c) — One oscillation in azimuth is not the Schuler period definition.
  • (d) — 63 minutes is not correct.
Instructor's Note: The Schuler period of 84 minutes = same as the orbital period of a satellite at Earth's surface. An INS aligned to this period will maintain a stable earth-vertical reference despite vehicle acceleration. This prevents the Schuler oscillation error.

Flight Instruments — Source Answer Key

Complete source answer key — Flight Instruments (Questions 1–164):
QAQAQAQAQAQA
1b2b3a4b5a6c
7b8a9c10b11b12c
13b14d15a16c17b18c
19b20b21b22a23d24d
25b26d27a28c29d30b
31d32c33d34a35c36b
37a38d39a40b41c42b
43c44b45c46a47d48c
49a50a51a52b53d54a
55a56a57b58a59a60d
61a62b63d64b65b66a
67d68b69a70a71a72c
73c74c75b76c77c78a
79c80c81b82c83c84a
85d86c87a88b89d90d
91b92d93a94a95b96c
97d98c99a100b101a102a
103c104b105c106c107d108d
109a110b111a112d113b114a
115c116a117c118a119c120c
121a122c123a124a125a126d
127c128c129a130c131d132c
133a134a135d136c137c138c
139c140a141d142b143b144b
145b146c147d148b149a150d
151c152a153b154a155a156b
157b158a159a160a161c162d
163a164d
Section B — Automatic Flight Questions
Coverage: Chapters 24–31 — FANS, Flight Director, Autopilot, Autoland, Autothrottle, Yaw Dampers, Control Laws, AFCS Revision. 58 questions total.

Key Questions with Detailed Explanations

B-Q5.At 50 feet AGL during an autoland, what happens to the glide slope signal?
  1. It continues to be actioned
  2. It is disconnected
  3. It is factored for range
  4. It is used to flare the aircraft
Correct Answer: (b) It is disconnected
Explanation: At 50 ft AGL during autoland, the glide slope signal is disconnected. Below this height, the flare is initiated using the radio altimeter signal. The GS signal is unreliable at very short ranges due to the extreme geometry (near vertical approach to the GS transmitter). The flare law takes over, using radio altimeter height to initiate the pitch-up and thrust reduction.
Why the other options are wrong:
  • (a) — The GS signal is not reliable at very low height and is purposely disconnected.
  • (c) — "Factored for range" is not correct; the signal is simply disconnected.
  • (d) — The radio altimeter (not GS) is used to flare the aircraft.
Instructor's Note: Autoland flare sequence: GS controls approach → at 50 ft, GS disconnected → radio altimeter triggers flare → at 15 ft, autothrottle retards to idle (flare complete) → touchdown via roll-out mode.
B-Q42.If only a single A/P is used to climb, cruise and approach, following a failure:
  1. it is fail-passive with redundancy
  2. it is fail-operational and will not disconnect
  3. it is fail-soft and will not disconnect
  4. it is fail-safe and will disconnect
Correct Answer: (d) it is fail-safe and will disconnect
Explanation: A single autopilot system, when it fails, must disconnect safely — this is called fail-safe. It disconnects on failure, alerting the crew to take manual control. Fail-passive systems (used in autoland) also disconnect but in a neutral position. Fail-operational systems (dual/triple autopilots) can continue the approach after one failure. A single AP has no redundancy, so it must disconnect cleanly on failure.
Why the other options are wrong:
  • (a) — Fail-passive with redundancy applies to a duplex system used for CAT II approaches, not a single AP.
  • (b) — Fail-operational requires a triplex system; the single AP cannot continue operating after failure.
  • (c) — Fail-soft means the system degrades gradually but does not apply to a single AP.
Instructor's Note: Single AP = fail-safe (disconnects). Duplex AP = fail-passive (disconnects in neutral → CAT II). Triplex AP = fail-operational (continues after 1 failure → CAT III). These three categories and their autoland capability levels are a core AFCS exam topic.

Automatic Flight — Source Answer Key

QAQAQAQAQAQA
1a2a3b4b5b6c
7a8b9a10d11a12b
13d14d15a16d17a18a
19d20c21d22b23d24a
25d26c27b28b29c30b
31b32c33b34d35b36a
37d38b39c40a41a42d
43d44a45b46b47c48b
49c50d51b52a53a54d
55a56c57d58b
Section C — Warning & Recording Questions
Coverage: Chapters 32–37 — Flight Warning Systems, Aerodynamic Warnings (stall, GPWS), GPWS modes, TCAS, FDR, CVR. 33 questions total.

Key Questions with Detailed Explanations

C-Q1.The primary input to a basic stall warning system is:
  1. angle of attack
  2. IAS
  3. slat/flap position
  4. MNO
Correct Answer: (a) angle of attack
Explanation: The primary input to a stall warning system is angle of attack (AoA). Stall is fundamentally an aerodynamic phenomenon that occurs at a critical AoA, not at a specific IAS. The AoA sensor (vane or probe on the fuselage) provides the primary signal. Configuration (flap/slat position) adjusts the stall threshold but is secondary. Note that a basic stall warning (e.g., a stall warning horn driven by AoA alone) uses only AoA.
Why the other options are wrong:
  • (b) — IAS is not the primary stall warning input; stall occurs at a given AoA regardless of weight.
  • (c) — Flap/slat position modifies the stall AoA threshold — it is a secondary input, not the primary.
  • (d) — Mach number (MNO) is not a primary stall warning input.
Instructor's Note: AoA = primary stall warning input. Configuration adjusts the threshold. Weight/bank angle are factored in more sophisticated systems. The vane-type AoA sensor on the fuselage side is the key sensor.
C-Q3.An FDR fitted to an aircraft of over 5,700 kg after April 1998 must record for:
  1. 10 hours
  2. 25 hours
  3. 30 minutes
  4. 60 minutes
Correct Answer: (b) 25 hours
Explanation: Aircraft above 5,700 kg MTOM require an FDR with a minimum recording time of 25 hours. Registration after 1 April 1998 makes this Case 1. The 10-hour recording time applies to turbine-powered aircraft below 5,700 kg MTOM but with more than 9 passenger seats registered after 1 April 1998.
Why the other options are wrong:
  • (a) — 10 hours applies to aircraft below 5,700 kg MTOM.
  • (c) — 30 minutes is the minimum CVR recording time (not FDR).
  • (d) — 60 minutes is not a standard FDR recording time threshold.
Instructor's Note: FDR recording: >5,700 kg = 25 hrs. <5,700 kg (turbine, >9 seats) = 10 hrs. CVR recording: standard = 30 min; >5,700 kg post-Apr '98 = 2 hrs. These four figures are often cross-tested.

Warning & Recording — Source Answer Key

QAQAQAQAQAQA
1a2a3b4a5d6c
7a8c9d10a11c12d
13b14b15a16c17a18d
19a20a21a22b23c24b
25b26b27b28c29c30b
31a32c33b
Section D — Engine Instruments Questions
Coverage: Chapters 38–39 — Engine instrumentation (EPR, torque, tacho, EGT, pressure gauges, vibration, fuel gauges) and electronic instrumentation (EICAS, ECAM). 52 questions total.

Key Questions with Detailed Explanations

D-Q1.What type of sensor is used to measure the output of a low-pressure booster pump?
  1. Bourdon tube
  2. Aneroid capsule
  3. Bellows
  4. Differential capsule
Correct Answer: (c) Bellows
Explanation: The Oxford text explicitly states: "The bellows type element... is typically used to measure pressures like the aircraft's LP booster pump output." Bellows can handle variable pressure ranges (high, low, or differential) and are specifically cited for LP booster pump pressure measurement.
Why the other options are wrong:
  • (a) — Bourdon tube = high pressure (e.g., engine oil pressure).
  • (b) — Aneroid capsule = low pressure but specifically sealed/evacuated; used in altimeters for ambient pressure sensing.
  • (d) — A differential capsule measures differential pressure; not specifically cited for LP booster pump.
Instructor's Note: LP booster pump = bellows. Oil pressure = Bourdon tube. Altimeter = aneroid capsule. These specific applications are memorisable pairings for the exam.
D-Q10.During the take-off run, the effect of increasing airspeed is to cause the EPR indication to:
  1. increase, due to ram rise
  2. fall, due to increase of intake pressure relative to jet pipe pressure
  3. remain constant as the jet pipe and intake pressures increase at the same rate
  4. fall, then gradually return to the original setting as V2 is reached
Correct Answer: (b) fall, due to increase of intake pressure relative to jet pipe pressure
Explanation: As airspeed increases during the take-off roll, the engine intake pressure increases due to the ram effect (more dynamic pressure). The jet pipe pressure initially does not increase proportionally at low airspeed. Therefore EPR (= jet pipe pressure ÷ intake pressure) falls. This is the well-known EPR apparent drop. Standard procedure: set EPR before 60 kt, do not increase power after that speed.
Why the other options are wrong:
  • (a) — EPR falls (not increases) as speed increases, because intake pressure rises faster than jet pipe pressure at low speed.
  • (c) — The two pressures do NOT increase at the same rate at low airspeed.
  • (d) — The EPR falls and does not return until airspeed exceeds V2, when the increase in intake pressure is passed through the engine to the jet pipe.
Instructor's Note: EPR drop on T/O is an APPARENT fall — not a real thrust loss. The standard answer is: set EPR before 60 kt, then do not touch throttles. After V2, as speed increases further, the ram pressure passes through to the jet pipe and EPR returns to the set value.
D-Q52.EPR is a:
  1. ratio between ambient pressure and exhaust pressure
  2. ratio between ambient pressure and fan pressure
  3. ratio between intake pressure and compressor delivery pressure
  4. ratio between exhaust pressure and intake pressure
Correct Answer: (d) ratio between exhaust (jet pipe) pressure and intake pressure
Explanation: EPR = Engine Pressure Ratio = jet pipe pressure (P7) ÷ engine air intake pressure (P1). A higher EPR indicates more thrust. The numerator is the exhaust/jet pipe pressure and the denominator is the intake pressure.
Why the other options are wrong:
  • (a) — Ambient pressure = intake pressure in still air, but the ratio is exhaust ÷ intake (not ambient ÷ exhaust).
  • (b) — Fan pressure is part of the integrated EPR on some turbofans, not the basic EPR definition.
  • (c) — Intake ÷ compressor delivery is a compressor pressure ratio, not EPR.
Instructor's Note: EPR = P7 (jet pipe) / P1 (intake). A ratio >1 means engine is producing thrust. The greater the EPR, the higher the thrust. This fundamental definition should be memorised verbatim.

Engine Instruments — Source Answer Key

QAQAQAQAQAQA
1c2b3b4d5b6c
7b8d9c10c11d12d
13d14b15b16d17b18d
19a20d21d22a23a24b
25d26d27c28a29b30d
31d32b33c34b35a36b
37b38a39a40b41c42d
43d44b45c46a47c48d
49a50d51a52d
Section E — Specimen Questions
Coverage: Mixed topics across the full Instrumentation syllabus. These questions include source explanations. 55 questions total — key questions with full worked explanations below.
E-Q1.A modern radio altimeter uses the frequency band:
  1. VHF — 30–300 MHz
  2. SHF — 3,000 MHz–30 GHz
  3. UHF — 300 MHz–3 GHz
  4. HF — 3 MHz–30 MHz
Correct Answer: (b) SHF — 3,000 MHz–30 GHz
Explanation: Radio altimeters (radar altimeters) operate in the SHF (Super High Frequency) band, typically around 4.2–4.4 GHz. This frequency provides the required short wavelength for accurate height measurement at low altitudes (0–2,500 ft). FM-CW (Frequency Modulated Continuous Wave) technique is used.
Why the other options are wrong:
  • (a) — VHF (30–300 MHz) is used for VOR, localiser, communications — not radio altimeters.
  • (c) — UHF (300 MHz–3 GHz) is used for DME, transponders — not radio altimeters.
  • (d) — HF (3–30 MHz) is used for long-range communications — not radio altimeters.
Instructor's Note: Radio altimeter = SHF (~4.3 GHz). This is in the microwave/centimetric wavelength range. The FM-CW technique allows measurement of height from 0 ft to 2,500 ft AGL with high accuracy.
E-Q8.Which of the following are modes of the GPWS?
  1. Excessive sink rate
  2. Altitude loss after take-off or go-around
  3. Excessive glide slope deviation
  4. High climb rate
  5. Flaps in the incorrect position
  6. High altitude descent
  7. Stall
  1. i, ii, iii, v
  2. ii, iii, v, vii
  3. i, ii, iii, vii
  4. iii, iv, v, vi
Correct Answer: (a) i, ii, iii, v — Excessive sink rate; altitude loss after T/O; excessive GS deviation; flaps incorrect
Explanation: The six basic GPWS modes are: Mode 1 = Excessive descent rate (i); Mode 2 = Excessive terrain closure; Mode 3 = Altitude loss after T/O or go-around (ii); Mode 4 = Proximity not configured (unsafe gear/flap — v); Mode 5 = Below glide slope (iii, noting this is below GS, not excessive deviation above); Mode 6 = Altitude call-outs. Option (a) matches three confirmed modes. "High climb rate" (iv), "high altitude descent" (vi), and "stall" (vii) are NOT GPWS modes.
Why the other options are wrong:
  • (b) — Stall (vii) is not a GPWS mode.
  • (c) — Stall (vii) is not a GPWS mode; and this combination is not standard.
  • (d) — High climb rate (iv) and high altitude descent (vi) are not GPWS modes.
Instructor's Note: Six GPWS modes: 1=sink rate, 2=terrain closure, 3=altitude loss T/O, 4=proximity not configured (gear/flap), 5=below GS, 6=callouts. TAWS/EGPWS adds predictive terrain capability but the 6 basic modes remain the examination foundation.
E-Q9.An aircraft is travelling at 120 kt. What angle of bank would be required for a rate-one turn?
  1. 30°
  2. 12°
  3. 19°
  4. 35°
Correct Answer: (c) 19°
Explanation: Rate 1 turn = 3°/second (360° in 2 minutes). Bank angle for rate 1 ≈ TAS/10 + 7. At 120 kt TAS: 120/10 + 7 = 12 + 7 = 19°. More precisely: tan(bank) = (rate × TAS) / (g × 180/π). Rate = 3°/s = 0.0524 rad/s; TAS = 120 kt = 202 ft/s. tan(bank) = (0.0524 × 202) / 32.2 = 10.58/32.2 = 0.329 → bank = arctan(0.329) ≈ 18.2° ≈ 19°.
Why the other options are wrong:
  • (a) — 30° corresponds to a faster speed (approximately 230 kt for rate 1).
  • (b) — 12° ≈ bank for rate 1 at 50 kt.
  • (d) — 35° ≈ bank for rate 1 at approximately 280 kt.
Instructor's Note: Quick formula: Rate 1 bank angle ≈ TAS/10 + 7. At 120 kt: 12+7=19°. At 180 kt: 18+7=25°. At 240 kt: 24+7=31°. This formula is approximate but accurate to within 1–2° for typical airspeeds.
E-Q10.An aircraft is travelling at 100 kt forward speed on a 3° glide slope. What is its rate of descent?
  1. 500 ft/min
  2. 300 ft/min
  3. 250 ft/min
  4. 500 ft/sec
Correct Answer: (a) 500 ft/min
Explanation: Standard formula: Rate of descent = groundspeed (kt) × 5 × glide slope angle (°). At 100 kt and 3°: ROD = 100 × 5 × 3 / 3 = 500 ft/min. More precisely: ROD = GS × tan(3°) × 101.3 ft/min per kt ≈ 100 × 0.0524 × 101.3 ≈ 531 ft/min, but the standard answer using the 5× rule gives 500. The rule-of-thumb: ROD = GS × (GPA × ~5/3) ≈ GS × 5 for 3° ≈ 500 ft/min at 100 kt.
Why the other options are wrong:
  • (b) — 300 ft/min is too low for this combination.
  • (c) — 250 ft/min is too low (would be correct for ~50 kt).
  • (d) — 500 ft/sec is a unit error (500 ft/min, not ft/sec).
Instructor's Note: Memorise: 3° GS at 150 kt ≈ 750 ft/min; at 120 kt ≈ 600 ft/min; at 100 kt ≈ 500 ft/min. These benchmarks cover the most common exam speeds. The general rule: ROD (ft/min) ≈ 5 × GS (kt) for a 3° slope.

Specimen Questions — Source Answer Key (E-Q1 to E-Q55)

QAQAQAQAQAQA
1b2b3b4b5c6c
7d8a9c10a11b12a
13b14a15c16a17b18a
19a20d21a22c23c24b
25a26a27b28b29c30c
31c32b33d34d35b36a
37c38b39a40c41b42c
43c44a45d46a47c48b
49a50d51c52a53a54d
55c
Section F — Specimen Examination Paper
Instructions: This is a timed specimen exam paper — 55 questions. Attempt all questions before checking answers. Each question carries the marks weight indicated (w1 = 1 mark, w2 = 2 marks, w3 = 3 marks). Allow approximately 75 minutes for this paper.

Selected Key Questions with Explanations

F-Q11.What correction is given by TCAS?
  1. Turn left or right
  2. Climb or descend
  3. Contact ATC on receipt of a resolution advisory
  4. Climb or descend at 500 ft/min
Correct Answer: (b) Climb or descend
Explanation: TCAS II provides Resolution Advisories (RAs) that give vertical manoeuvre guidance only — it tells the pilot to climb or descend (and at what rate). TCAS does NOT give horizontal manoeuvre guidance (turn left/right). The vertical-only guidance is why TCAS RAs specify climb/descend rates (e.g., 1,500 or 2,500 ft/min) rather than headings.
Why the other options are wrong:
  • (a) — TCAS does NOT give lateral (turn) manoeuvre instructions — vertical only.
  • (c) — On receipt of a TCAS RA, the crew must follow the RA, NOT contact ATC first. ATC instructions are secondary to TCAS RAs.
  • (d) — TCAS specifies minimum vertical rates but does not always specify exactly 500 ft/min; the RA specifies a rate range depending on the threat geometry.
Instructor's Note: TCAS RA = vertical guidance only (climb or descend). The correct response: smoothly and immediately comply with the RA. Then notify ATC. NEVER follow ATC instructions that conflict with an active RA. This is the critical safety rule.

Specimen Examination Paper — Source Answer Key

QAWtQAWtQAWtQAWt
1cw12aw13bw14aw1
5cw16bw17aw18aw1
9cw110aw111cw112bw2
13cw114cw115aw116dw2
17dw118cw119aw120aw1
21bw222aw223dw124cw3
25cw226dw127bw128dw1
29aw130dw131bw232bw1
33dw134bw135aw136cw1
37dw138bw139cw140dw1
41aw142aw243bw144dw1
45cw146bw247cw148dw2
49bw150cw151dw152dw1
53cw154cw155bw1

Master Answer Keys — All Sections

Exam Strategy for Chapter 40:
Top 10 Most-Tested Numerical Values — Final Revision:
#ValueTopic
184 minutesSchuler period (INS)
225 hoursFDR minimum recording time (>5,700 kg)
330 minutesCVR minimum recording time (standard)
42 hoursCVR minimum recording time (>5,700 kg, post Apr 1998)
55,700 kgMTOM threshold — FDR/CVR requirement
672 hoursMax time since u/s — dispatch with inoperative FDR/CVR
78 flightsMax consecutive flights — dispatch with inoperative FDR/CVR
860 ktEPR must be set before this speed on T/O roll
950 ftAGL — GS signal disconnected during autoland flare
102,500 ftMaximum operating height of GPWS
Capt. Pankaj Pahil