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RTR(A) — CH.12

Navigation AidsRadio Telephony — DGCA CPL practice questions

Question 1 of 20

The ADF resolves the 180° ambiguity of the loop antenna by:

A.Using a higher frequency
B.Adding a sense antenna to form a cardioid with a single null
C.Rotating the aircraft
D.Comparing 90 and 150 Hz tones

All 20 questions — Navigation Aids

Radio Telephony · DGCA CPL. The correct option is marked on each.

  1. Q1. The ADF resolves the 180° ambiguity of the loop antenna by:

    • A.Using a higher frequency
    • B.Adding a sense antenna to form a cardioid with a single null
    • C.Rotating the aircraft
    • D.Comparing 90 and 150 Hz tones

    Why: dding a sense antenna to form a cardioid with a single null. The omnidirectional sense antenna combines with the figure-of-eight loop to give a cardioid with one null.

  2. Q2. Heading 350°(M), ADF relative bearing 040°. The QDM is:

    • A.310°
    • B.030°
    • C.050°
    • D.210°

    Why: QDM = RB + HDG = 040 + 350 = 390 − 360 = 030°.

  3. Q3. A VOR's reference and variable 30 Hz signals are in phase when the aircraft is:

    • A.Overhead the station
    • B.Due magnetic north of the station
    • C.On the 180° radial
    • D.At the cone of confusion

    Why: ue magnetic north of the station. By design the phase difference is zero on the 360° (north) radial; it then equals the radial.

  4. Q4. In a conventional VOR, the reference phase signal is carried as:

    • A.30 Hz AM
    • B.30 Hz FM on a 9960 Hz sub-carrier
    • C.A 90 Hz tone
    • D.A Morse code

    Why: CVOR: reference = FM on 9960 Hz sub-carrier (same in all directions); variable = AM from the rotating pattern.

  5. Q5. The ILS localizer indicates centreline when:

    • A.90 Hz dominates
    • B.150 Hz dominates
    • C.The depth of modulation of 90 and 150 Hz is equal (DDM = 0)
    • D.The DME reads zero

    Why: On the centreline the two tones are received equally — zero difference in depth of modulation.

  6. Q6. ILS marker beacons all transmit on:

    • A.329–335 MHz
    • B.108.10–111.95 MHz
    • C.75 MHz
    • D.1030 MHz

    Why: Outer/middle/inner markers share a 75 MHz carrier, differing in tone and code.

  7. Q7. A false glide slope is typically encountered:

    • A.Below the true 3° path
    • B.Above the true path (e.g. ~6°), if intercepted from above
    • C.On the localizer back course
    • D.Only at night

    Why: bove the true path (e.g. ~6°), if intercepted from above. Antenna side-lobes create false slopes at higher angles; intercept the GP from below to avoid them.

  8. Q8. DME measures distance by:

    • A.Comparing two 30 Hz phases
    • B.Timing the round trip of pulse pairs, less a 50 µs ground delay
    • C.Measuring received signal strength
    • D.Counting satellites

    Why: The interrogator times the reply (minus the fixed 50 µs transponder delay) to derive slant range.

  9. Q9. At 12,000 ft (2 NM) directly overhead a DME, the indicator reads approximately:

    • A.0 NM
    • B.2 NM
    • C.6 NM
    • D.12 NM

    Why: Overhead, ground range is zero so slant range equals height: 12,000 ft ≈ 2 NM.

  10. Q10. An SSR transponder is interrogated on, and replies on, respectively:

    • A.1090 / 1030 MHz
    • B.1030 / 1090 MHz
    • C.329 / 335 MHz
    • D.75 / 75 MHz

    Why: Ground interrogates on 1030 MHz; aircraft replies on 1090 MHz.

  11. Q11. Side-Lobe Suppression in SSR prevents:

    • A.Garbling
    • B.FRUIT
    • C.Replies to the antenna's side lobes giving false bearings
    • D.Altitude errors

    Why: A P2 control pulse stops the transponder replying when interrogated via a side lobe.

  12. Q12. The transponder code for radio communications failure is:

    • A.7500
    • B.7600
    • C.7700
    • D.7000

    Why: 7500 hijack, 7600 radio failure, 7700 emergency.

  13. Q13. Which radar requires equipment carried aboard the aircraft to function?

    • A.Primary radar
    • B.Secondary radar
    • C.Both
    • D.Neither

    Why: SSR needs a transponder; primary radar works on the passive echo from any target.

  14. Q14. A 3-D GNSS position fix requires a minimum of how many satellites?

    • A.2
    • B.3
    • C.4
    • D.6

    Why: Three for position plus one to solve the receiver clock error = four.

  15. Q15. The GPS L1 carrier frequency is:

    • A.1227.60 MHz
    • B.1575.42 MHz
    • C.1090 MHz
    • D.1176.45 MHz

    Why: L1 = 1575.42 MHz (L2 = 1227.6, L5 = 1176.45).

  16. Q16. GAGAN is best described as:

    • A.A ground-based primary radar
    • B.A satellite-based augmentation system (SBAS) for the Indian region
    • C.India's own satellite constellation for global coverage
    • D.An inertial system

    Why: satellite-based augmentation system (SBAS) for the Indian region. GAGAN broadcasts corrections + integrity over India via GEO satellites. (NavIC/IRNSS is the regional constellation.)

  17. Q17. The essential difference between RNAV and RNP is that RNP adds:

    • A.A higher cruising altitude
    • B.On-board performance monitoring and alerting
    • C.A second VOR
    • D.Mode S

    Why: RNP = RNAV plus the requirement that the system monitors its own accuracy and alerts the crew.

  18. Q18. A TCAS Resolution Advisory commands a manoeuvre in:

    • A.The horizontal plane (turn)
    • B.The vertical plane (climb/descend)
    • C.Any plane
    • D.Speed only

    Why: RAs are vertical-only; never turn in response to an RA.

  19. Q19. The radio altimeter operates around 4200–4400 MHz and indicates:

    • A.Pressure altitude
    • B.True height above the surface directly below (AGL)
    • C.Altitude above mean sea level
    • D.Slant range to a beacon

    Why: FM-CW radio altimeter gives AGL, typically 0–2500 ft.

  20. Q20. A characteristic of INS used alone is that its position error:

    • A.Is constant regardless of time
    • B.Grows with time (drift), typically a few NM per hour
    • C.Depends on satellite geometry
    • D.Only occurs over the sea

    Why: Integration accumulates error, so unaided INS drifts (~2 NM/hr); GNSS is used to bound it.