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Free ATPL principles of flight 081 questions: 9 with answers

9 principles of flight questions from the ATPL bank of ForAllPilots, with no account and no sign-up. Answer in your head, then open the answer: the right letter and its explanation are under each question. The full bank has 1,335 questions for this subject.

The 9 questions

  1. Question 1 · 081 01 Subsonic aerodynamics

    In steady level flight, CL is 0.86 at 180 kt. If speed changes to 190 kt with weight unchanged, the new CL is approximately:

    • A0.46
    • B0.86
    • C1.08
    • D0.77
    Show the answer+

    Answer: D. 0.77

    In level flight lift equals weight, so CL is inversely proportional to V². CL2 = 0.86 x (180 / 190)² = 0.86 x 0.897 = about 0.77.

  2. Question 2 · 081 01 Subsonic aerodynamics

    Compared with the flap up configuration the maximum angle of attack for the flaps down configuration is

    • Alarger.
    • Bunchanged.
    • Csmaller or larger depending on flap deflection.
    • Dsmaller.
    Show the answer+

    Answer: D. smaller.

    Trailing edge flaps increase camber and CLmax, but the flow separates at a lower angle of attack. The critical angle of attack with flaps down is therefore smaller than with flaps up.

  3. Question 3 · 081 01 Subsonic aerodynamics

    An aeroplane accelerates from 80 kt to 160 kt at a load factor equal to 1. The induced drag coefficient (i) and the induced drag (ii) alter with the following factors:

    • A(i) 1/2 (ii) 1/16
    • B(i) 1/4 (ii) 2
    • C(i) 4 (ii) 1/2
    • D(i) 1/16 (ii) 1/4
    Show the answer+

    Answer: D. (i) 1/16 (ii) 1/4

    In level flight CL is proportional to 1/V², and the induced drag coefficient to CL², so to 1/V⁴. Doubling the speed divides CDi by 16. Induced drag = CDi x ½ρV²S is proportional to 1/V², so it is divided by 4.

  4. Question 4 · 081 02 High-speed aerodynamics

    Climbing at a constant Mach Number up to FL 350 the TAS will:

    • Aincrease.
    • Bdecrease.
    • Cremain constant.
    • Dfirst increase, then decrease.
    Show the answer+

    Answer: B. decrease.

    Climbing at constant Mach with falling temperature (up to the tropopause), the local speed of sound decreases, so TAS (=M×a) decreases too.

  5. Question 5 · 081 03 stall and upset prevention

    Compared with level flight prior to the stall, the lift (1) and drag (2) in the stall change as follows :

    • A(1) decreases (2) decreases.
    • B(1) decreases (2) increases.
    • C(1) increases (2) decreases.
    • D(1) increases (2) increases.
    Show the answer+

    Answer: B. (1) decreases (2) increases.

    At the stall the flow separates from the upper surface: lift falls and drag (mostly pressure drag) rises sharply compared with level flight before the stall.

  6. Question 6 · 081 04 Stability

    An aircraft is said to be statically stable if, following an increase in angle of attack, the gradients ΔCm/ΔCL or ΔCm/Δα are:

    • Anegative
    • Bpositive
    • Czero
    • Dequal to Cm0
    Show the answer+

    Answer: A. negative

    An aeroplane is statically stable in pitch if an increase in angle of attack produces a nose-down pitching moment, which opposes the disturbance. This means the slope dCm/dCL (or dCm/dα) is negative.

  7. Question 7 · 081 06 Limitations

    VMO :

    • Ais equal to the design speed for maximum gust intensity.
    • Bshould be chosen in between VC and VD
    • Cshould be not greater than VC.
    • Dis the calibrated airspeed at which MMO is reached at 35 000 ft.
    Show the answer+

    Answer: C. should be not greater than VC.

    Under CS 25.1505, the maximum operating limit speed VMO may not exceed the design cruising speed VC, and must be sufficiently below VD.

  8. Question 8 · 081 06 Limitations

    An aeroplane has VS1g=130 kt and a positive limit load factor of 2.5g. The approximate design manoeuvring speed VA is:

    • A144 kt
    • B130 kt
    • C268 kt
    • D206 kt
    Show the answer+

    Answer: D. 206 kt

    The design manoeuvring speed is about VS x square root of the limit load factor: 130 x square root of 2.5 = 130 x 1.58 = about 206 kt.

  9. Question 9 · 081 08 Flight mechanics

    An aeroplane in a coordinated turn at TAS 170 kt and 20° bank angle has an approximate rate of turn of:

    • A3.28 °/sec
    • B3.34 °/sec
    • C1.4 °/sec
    • D2.34 °/sec
    Show the answer+

    Answer: D. 2.34 °/sec

    Rate of turn (°/s) = 1,091 x tan bank / TAS (kt) = 1,091 x tan 20° / 170 = 1,091 x 0.364 / 170 = about 2.34°/s.

Each answer was checked again before publication; like every explanation on the site, these may contain errors. Some explanations were rewritten for this page, sometimes with the help of an AI. Spotted a mistake? Write to us at contact@forallpilots.com.

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