DGCA CPL Air Navigation — question bank and syllabus

Why that is the answer

In an INS operating in NAV mode, the east/west accelerometer senses acceleration along the local east-west (parallel of latitude) axis. Integrating this acceleration once with respect to time yields velocity — specifically velocity along the local parallel of latitude (departure rate). A second integration of that velocity would then yield departure (distance travelled east-west), from which change of longitude and hence vehicle longitude are subsequently derived using the local radius of the parallel. Option (b) 'departure' is the result of the SECOND integration, not the first, so it is wrong. Options (a) and (d), longitude and change of longitude, are downstream computed quantities requiring further conversion using latitude, not a direct integration result. Only option (c) correctly identifies the first integration's output as velocity along the parallel of latitude.

Why that is the answer

In a conventional VOR, the reference phase signal is frequency-modulated (FM) onto a 9960 Hz subcarrier while the variable phase signal is amplitude-modulated (AM) at 30 Hz on the main carrier. In a DVOR, this modulation roles are reversed: the reference phase signal is amplitude-modulated at 30 Hz on the main carrier (radiated omnidirectionally from a central antenna), while the variable phase signal is effectively frequency-modulated at 9960 Hz produced by the sideband antennas being electronically commutated around the circular array (simulating rotation via a Doppler-shifted signal). This role-reversal is precisely why option (b) is correct, and option (a), claiming the same modulation roles as VOR, is incorrect and describes conventional VOR, not DVOR.

Why that is the answer

Resolving the TAS of 235 kt along the true heading of 076° into north/east components, and separately resolving the given wind (040°/40kt, blowing toward 220°) into its own components, allows the groundspeed vector to be built as their vector sum (GS = TAS + W/V). Adding the components gives a resultant groundspeed of about 204 kt on a track close to 083°T. Since the track (083°) is numerically greater than the heading (076°), the aircraft has drifted to the right of its heading, giving a drift of about 7° right. Option (a) 5R−207kt and (d) 5L−255kt use an incorrect drift magnitude and sign respectively, while (c) 7L−269kt gets the drift magnitude right but the sign and resulting groundspeed wrong, effectively treating the wind as a tailwind pushing the track left instead of right.

1. Resolve TAS (235 kt) along heading 076°T into N/E components
2. Resolve W/V (040°/40kt, blowing toward 220°) into N/E components
3. Add vectors: GS vector = TAS vector + Wind vector
4. Magnitude of GS vector ≈ 204 kt; track direction ≈ 083°T
5. Drift = Track − Heading = 083° − 076° ≈ 7° Right
6. Result: 7° Right drift, GS ≈ 204 kt

Why that is the answer

Secondary Surveillance Radar operates on a fixed pair of frequencies defined by ICAO Annex 10: the ground interrogator transmits its interrogation pulses on 1030 MHz, and the airborne transponder replies on 1090 MHz. This separation prevents the interrogation and reply signals from interfering with each other and allows all SSR ground stations worldwide, plus ACAS/TCAS airborne equipment, to use a common frequency pair. Option (a) incorrectly shows both transmit and receive as 1030 MHz, which would mean the ground station receives its own interrogation frequency rather than the transponder's reply; (c) reverses the correct assignment; and (d) incorrectly shows both as 1090 MHz. Only option (b), transmitter 1030 MHz and receiver 1090 MHz, matches the standardised SSR frequency plan.