Forces in steady level flight?
Lift balances weight; thrust balances drag.
- Lift increases with density, speed squared, wing area and lift coefficient
- Same form for drag with CD
181 practice questions on Aerodynamics & Performance with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.
Use this page to revise Aerodynamics & Performance for the DGCA Technical General paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 181 questions in the TrueHeading bank for this topic.
In short: Lift balances weight; thrust balances drag.
Open the full set in the student zone to answer every question, track your accuracy and take timed mock tests.
14 ideas to know cold.
Lift balances weight; thrust balances drag.
The wing exceeds its critical angle of attack (about 15–16°) and the airflow separates.
Parasite drag grows with the square of speed. Induced drag falls with the square of speed.
Lift must increase, so load factor rises with bank angle.
Both depend on speed and bank angle.
Within about one wingspan of the ground the downwash is restricted. Induced drag falls and lift rises.
Trailing edge flaps raise CL max and drag; leading edge slats delay the stall to a higher angle of attack.
Sweepback raises the critical Mach number.
Vortices are strongest behind heavy, slow, clean aircraft.
Related to drag and power curves.
L = ½ ρ V² S CL.
Weight (↑), load factor (↑), flap (↓ with flaps), CG forward (↑), altitude (TAS ↑, IAS same), ice (↑).
Induced drag (from lift) and parasite drag (form, skin friction, interference). Wave drag at high Mach.
L/D at the best angle of attack: height lost × L/D = distance.
Tap “Show answer” after you have tried each one.
Answer: C. 1.31 g
n = 1 ÷ cos 40° = 1.31.
Answer: D. 112 kt
Vs(turn) = Vs × √n = 90 × √1.56 = 112 kt.
Answer: B. 75 kt
Vs(turn) = Vs × √n = 70 × √1.15 = 75 kt.
Answer: D. 463 kt
LSS = 38.95 × √(T in K) = 38.95 × √233 = 594 kt; TAS = M × LSS = 463 kt.
Answer: A. 667 kt
LSS = 38.95 × √293.1 = 667 kt.
Answer: A. 39°
tan(bank) = V × ω ÷ g = 149.2 × 0.0524 ÷ 9.81, so bank ≈ 39° (rule of thumb: TAS/10 + 7).
Answer: B. 520 ft/min
ROC = gradient × GS × 101.3 = 0.03 × 170 × 101.3 ≈ 516 ft/min.
Answer: B. 97.2 kN
L = ½ ρ V² S CL.
Answer: B. 171
Vs increases by √n = √(1/cos φ).
Answer: C. 12 NM
Distance = height × L/D (1 NM ≈ 6 076 ft).
Answer: A. increases with the square of airspeed
Parasite drag ∝ V².
Answer: D. weight divided by wing area
W/S, typically in kg/m² or N/m².
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