DGCA CPL Technical Specific — question bank and syllabus

Why that is the answer

Maximum traffic load equals the maximum permissible zero fuel mass minus the Dry Operating Mass, where the permissible zero fuel mass itself is the lowest of three constraints: the structural MZFM, the ZFM implied by the performance-limited landing mass minus the fuel remaining after landing, and the ZFM implied by the performance-limited take-off mass minus the total fuel carried at take-off. Fuel on board at take-off is trip fuel plus contingency/final reserve fuel plus alternate fuel = 12500+2300+1700 = 16500 kg; fuel remaining at landing (after trip fuel is burned) = 16500-12500 = 4000 kg. The landing-derived ZFM limit is 72250-4000 = 68250 kg; the take-off-derived ZFM limit is 92750-16500 = 76250 kg; the structural MZFM is 67530 kg, the lowest of the three, so it governs. Traffic load = 67530-60520 = 7010 kg, matching option a.

1. Given: DOM=60520 kg, Perf-limited TOM=92750 kg, Perf-limited LM=72250 kg, MZFM=67530 kg, Trip fuel=12500 kg, Contingency+final reserve=2300 kg, Alternate fuel=1700 kg
2. Fuel on board at take-off = 12500 + 2300 + 1700 = 16500 kg
3. Fuel remaining at landing = 16500 - 12500 = 4000 kg
4. Landing-derived ZFM limit = 72250 - 4000 = 68250 kg; Take-off-derived ZFM limit = 92750 - 16500 = 76250 kg
5. Governing (lowest) ZFM limit = min(67530, 68250, 76250) = 67530 kg (structural MZFM)
6. Maximum traffic load = ZFM limit - DOM = 67530 - 60520 = 7010 kg

Why that is the answer

If a pilot mistakenly selects a V1 higher than the correct value, and an engine fails at a speed just above this incorrect (too-high) V1, the crew will continue the take-off (since the decision speed has technically been reached/passed) even though the actual, correct V1 for continuing safely was lower. This means the take-off continues with less speed margin built into the certified performance calculations than intended, and the achieved climb-out speed profile — including V2 attainment — may be compromised: V2 may effectively be reached under conditions where climb performance is degraded relative to what was calculated, because the balanced field/climb performance data assumed the correct (lower) V1. The stop distance issue would arise from an engine failure recognized too late relative to accelerate-stop calculations if V1 were too LOW, not too high; over-rotation and rejecting-related stopping distance concerns are not the direct consequence described. The core problem with an artificially high V1 is degraded one-engine-out climb performance because the aeroplane continues the take-off later than the true safe V1, affecting the climb speed/gradient margins built around V2.

Why that is the answer

For a glider or power-off descent, the vertical and forward speed at a given angle of attack are both governed by the equilibrium glide speed, which itself is proportional to the square root of wing loading (weight divided by wing area). A heavier aeroplane must fly faster at the same angle of attack (same lift coefficient) to generate the extra lift needed to support its weight, so both its forward speed and its vertical (sink) speed at that angle of attack are larger than those of a lighter, otherwise identical aeroplane. Crucially, because lift-to-drag ratio (and hence best glide angle and glide distance) at a given angle of attack is unchanged by weight, glide distance capability is the same for both aircraft — ruling out (a) and (d). Option (c) is wrong because although glide ANGLE/RATIO is unaffected by mass, the actual speeds at a given AoA are not equal between different masses.

Why that is the answer

On short final in the landing configuration, the minimum approach/reference speed used is conventionally taken as 1.3 times the stalling speed in that configuration (VREF = 1.3 Vs), giving a standard 30% margin above the stall to protect against gust effects, manoeuvring loads and a safe flare margin. With a landing-configuration stalling speed of 100 kt, multiplying by 1.3 gives a minimum speed of 130 kt that the pilot must maintain on short final. The other options (115, 125, 120 kt) do not correspond to the standard 1.3 Vs margin and are simply incorrect multiples.

1. Given: VS (landing configuration) = 100 kt
2. Standard minimum approach speed margin = 1.3 x VS
3. Substitute: 1.3 x 100 kt
4. Result: 130 kt