VHF Direction Finder (VDF)Radio Navigation — DGCA CPL practice questions
Question 1 of 6
An aircraft has to communicate with a VHF station at a range of 300 NM. If the ground station is situated 2500 ft AMSL, which of the following is the lowest altitude at which contact is likely to be made?
All 6 questions — VHF Direction Finder (VDF)
Radio Navigation · DGCA CPL. The correct option is marked on each.
Q1. An aircraft has to communicate with a VHF station at a range of 300 NM. If the ground station is situated 2500 ft AMSL, which of the following is the lowest altitude at which contact is likely to be made?
- A.190 ft
- B.1,378 ft
- C.36,100 ft✓
- D.84,100 ft
Why: Working: 300 = 1.25 × (√2500 + √h) 300/1.25 = 240 = 50 + √h √h = 190 → h = 190² = 36,100 ft — The most common error is option (a) — computing √h = 190 and stopping there. Option (d) is the "addition instead of subtraction" trap. Always: isolate √h by subtracting the station term, then square to get h.
Q2. Class 'B' VHF DF bearings are accurate to within:
- A.± 1°
- B.± 5°✓
- C.± 2°
- D.± 10°
Why: Class B = ±5°. The four classes: A = ±2°, B = ±5°, C = ±10°, D = >±10°. — Most common wrong answer is (c) — swapping Class A (±2°) and Class B (±5°). Remember: A=2, B=5, C=10. Class B (±5°) is the normal best case in practice.
Q3. A VDF QDM given without an accuracy classification may be assumed to be accurate to within:
- A.2 degrees
- B.5 degrees✓
- C.7.5 degrees
- D.10 degrees
Why: The rule states: "Normally, bearings no better than Class B will be available." Therefore, when no accuracy class is specified, the bearing should be assumed to be Class B = ±5° . — This tests the "default assumption" rule. "Bearings no better than Class B" means Class B (±5°) is the default when no class is stated. This is directly examinable.
Q4. An aircraft at altitude 9000 ft wishes to communicate with a VHF/DF station situated at 400 ft AMSL. What is the maximum range at which contact is likely to be made?
- A.115 NM
- B.400 NM
- C.143 NM✓
- D.63.5 NM
Why: Working: Range = 1.25 × (√9000 + √400) = 1.25 × (94.87 + 20) = 1.25 × 114.87 = 143.6 ≈ 143 NM — √9000 ≈ 94.87 (use a calculator; this is not a round number). √400 = 20 exactly. Sum = 114.87. × 1.25 = 143.6 NM. Round to 143 NM.
Q5. An aircraft is passed a true bearing from a VDF station of 353°. If variation is 8°E and the bearing is classified as 'B' then the:
- A.QDM is 345° ± 5°
- B.QDR is 345° ± 2°
- C.QTE is 353° ± 5°✓
- D.QUJ is 353° ± 2°
Why: A "true bearing FROM the station" = QTE = 353° Class B accuracy = ±5° Therefore: QTE = 353° ± 5° Deriving all Q-codes for reference: QTE (True FROM) = 353° (given) QDR (Mag FROM) = 353° − 8°E variation = 345° QUJ (True TO) = 353° − 180° = 173° QDM (Mag TO) = 173° − 8° = 165° — This is the most complex question in the chapter. Three traps: Q-code identity, correct bearing value, correct accuracy class. Work through: (1) True bearing FROM = QTE; (2) Class B = ±5°. The most instructive wrong answer is (b) — it has the right bearing for QDR but the wrong class; a careful student should spot bot…
Q6. An aircraft at 19,000 ft wishes to communicate with a VDF station at 1400 ft AMSL. What is the maximum range at which contact is likely?
- A.175 NM
- B.400 NM
- C.62.5 NM
- D.219 NM✓
Why: Working: Range = 1.25 × (√19000 + √1400) = 1.25 × (137.84 + 37.42) = 1.25 × 175.26 = 219.1 ≈ 219 NM — Option (a) is the classic "forgot to multiply by 1.25" trap — the sum of square roots is 175.26, which exactly matches option (a). Always complete the full formula. √19000 ≈ 137.84; √1400 ≈ 37.42.