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CPL Test Series
Question Bank
Questions for Indigo gen NAV 4090-4604
Answer the following questions
Test Mode
Training Mode
1.
On a Transverse Mercator chart, scale is exactly correct along the
prime meridian and the equator
Equator, parallel of origin and prime vertical
meridian of tangency
datum meridian and meridian perpendicular to it
2.
When is the magnetic compass most effective?
About midway between the magnetic poles
On the geographic equator
In the region of the magnetic North Pole.
In the region of the magnetic South Pole.
3.
An aircraft travels from point A to point B, using the autopilot connected to the aircraft's inertial system. The coordinates of A (45°S 010°W) and B (45°S 030°W) have been entered.The true course of the aircraft on its arrival at B, to the nearest degree, is
284°
270°
277°
263°
4.
Approximately how many nautical miles correspond to 12 cm on a map with a scale of 1 : 2 000 000?
150
329
130
43
5.
Given:A polar stereographic chart whose grid is aligned with the zero meridian. Grid track 344°, Longitude 115°00'W,Calculate the true course?
099°
049°
279°
229°
6.
An aircraft departing A(N40º 00´ E080º 00´) flies a constant true track of 270º at a ground speed of 120 kt. What are the coordinates of the position reached in 6 HR?
N40º 00´ E068º 10´
N40º 00´ E060º 00´
N40º 00´ E064º 20´
N40º 00´ E070º 30´
7.
The following information is displayed on an Inertial Navigation System:GS 520 kt,True HDG 090°,Drift angle 5° right,TAS 480 kt.SAT (static air temperature) -51°C.The W/V being experienced is
225° / 60 kt
320° / 60 kt
220° / 60 kt
325° / 60 kt
8.
Given: Distance 'A' to 'B' is 475 NM,Planned GS 315 kt,ATD 1000 UTC,1040 UTC - fix obtained 190 NM along track.What GS must be maintained from the fix in order to achieve planned ETA at 'B'?
360 kt.
300 kt
320 kt.
340 kt
9.
Given: GS = 236 kt.Distance from A to B = 354 NMWhat is the time from A to B?
1 HR 09 MIN
1 HR 10 MIN
1 HR 30 MIN
1 HR 40 MIN
10.
An aircraft at FL370, M0.86, OAT -44°C, headwind component 110 kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. If the speed reduction were to be made 420 NM from the reporting point, what Mach Number is required?
M0.79
M0.75
M0.81
M0.73
11.
The angle between True North and Magnetic North is called
compass error
deviation
variation
drift
12.
What is the meaning of the term ""standard time"" ?
It is another term for UTC
It is the time zone system applicable only in the USA
It is the time set by the legal authorities for a country or part of a country
It is an expression for local mean time
13.
Given:Distance 'Q' to 'R' 1760 NMGroundspeed 'out' 435 ktGroundspeed 'back' 385 ktSafe endurance 9 HR The distance from 'Q' to the Point of Safe Return (PSR) between 'Q' and 'R' is
1313 NM
1467 NM
1642 NM
1838 NM
14.
At 47° North the chart distance between meridians 10° apart is 5 inches. The scale of the chart at 47° North approximates
1 : 8 000 000
1 : 2 500 000
1 : 6 000 000
1 : 3 000 000
15.
A ground feature appears 30° to the left of the centre line of the CRT of an airborne weather radar. If the heading of the aircraft is 355° (M) and the magnetic variation is 15° East, the true bearing of the aircraft from the feature is
130°
310°
160°
220°
16.
When turning right from 330°(C) to 040°(C) in the northern hemisphere, the reading of a direct reading magnetic compass will
under-indicate the turn and liquid swirl will decrease the effect
over-indicate the turn and liquid swirl will decrease the effect
over-indicate the turn and liquid swirl will increase the effect
under-indicate the turn and liquid swirl will increase the effect
17.
One of the errors inherent in a ring laser gyroscope occurs at low input rotation rates tending towards zero when a phenomenon known as 'lock-in' is experienced. What is the name of the technique, effected by means of a piezo-electric motor, that is used to correct this error?
dither
cavity rotation
beam lock
zero drop
18.
An aircraft is following a true track of 048° at a constant TAS of 210 kt. The wind velocity is 350° / 30 kt. The GS and drift angle are
200 kt, 3.5° right
192 kt, 7° right
225 kt, 7° left
192 kt, 7° left
19.
On a transverse Mercator chart, the scale is exactly correct along the
meridians of tangency
meridian of tangency and the parallel of latitude perpendicular to it
equator and parallel of origin
prime meridian and the equator
20.
For a landing on runway 23 (227° magnetic) surface W/V reported by the ATIS is 180/30kt. VAR is 13°E. Calculate the cross wind component?
20 kt
22 kt
15 kt
26 kt
21.
What is the time required to travel along the parallel of latitude 60° N between meridians 010° E and 030° W at a groundspeed of 480 kt?
1 HR 15 MIN
2 HR 30 MIN
5 HR 00 MIN
1 HR 45 MIN
22.
The azimuth gyro of an inertial unit has a drift of 0.01°/HR.After a flight of 12 HR with a ground speed of 500 kt, the error on the aeroplane position is approximately
6 NM
60 NM
12 NM
1 NM
23.
The automatic flight control system (AFCS) in an aircraft is coupled to the guidance outputs from an inertial navigation system (INS) and the aircraft is flying from waypoint No. 2 (60°00'S 070°00'W) to No. 3 (60°00'S 080°00'W).Comparing the initial track (°T) at 070°00'W and the final track (°T) at 080°00'W, the difference between them is that the initial track is approximately
9° less than the final one
5° greater than the final one
5° less than the final one
9° greater than the final one
24.
Given:ILS GP angle = 3.5 DEG,GS = 150 kt.What is the approximate rate of descent?
700 FT/MIN
1000 FT/MIN
900 FT/MIN
800 FT/MIN
25.
An aircraft flies a great circle track from 56° N 070° W to 62° N 110° E. The total distance travelled is?
5420 NM
2040 NM
1788 NM
3720 NM
26.
Which one of the following statements is correct concerning the appearance of great circles, with the exception of meridians, on a Polar Stereographic chart whose tangency is at the pole ?
Any straight line is a great circle
They are curves convex to the Pole
The higher the latitude the closer they approximate to a straight line
They are complex curves that can be convex and/or concave to the Pole
27.
The constant of the cone, on a Lambert chart where the convergence angle between longitudes 010°E and 030°W is 30°, is
0.50
0.75
0.64
0.40
28.
An aircraft at FL120, IAS 200kt, OAT -5° and wind component +30kt, is required to reduce speed in order to cross a reporting point 5 MIN later than planned. Assuming flight conditions do not change, when 100 NM from the reporting point IAS should be reduced to
165 kt
169 kt
159 kt
174 kt
29.
A negative (westerly) magnetic variation signifies that
True North is West of Magnetic North
True North is East of Magnetic North
Compass North is East of Magnetic North
Compass North is West of Magnetic North
30.
Given: Distance 'A' to 'B' is 100 NM,Fix obtained 40 NM along and 6 NM to the left of course.What heading alteration must be made to reach 'B'?
6° Right
9° Right
15° Right
18° Right
31.
Given: TAS = 132 kt,HDG (T) = 053°,W/V = 205/15kt.Calculate the Track (°T) and GS?
052 - 143 kt
057 - 144 kt
051 - 144 kt
050 - 145 kt
32.
Some inertial reference and navigation systems are known as ""strapdown"". This means that
only the gyros, and not the accelerometers, become part of the unit's fixture to the aircraft structure
gyros and accelerometers need satellite information input to obtain a vertical reference
gyros and accelerometers are mounted on a stabilised platform in the aircraft
the gyroscopes and accelerometers become part of the unit's fixture to the aircraft structure
33.
On the 27th of February, at 52°S and 040°E, the sunrise is at 0243 UTC. On the same day, at 52°S and 035°W, the sunrise is at
2143 UTC
0243 UTC
0743 UTC
0523 UTC
34.
Which of the following statements concerning earth magnetism is completely correct?
An isogonal is a line which connects places of equal dip, the aclinic is the line of zero magnetic dip
An isogonal is a line which connects places with the same magnetic variation, the aclinic connects places with the same magnetic field strength
An isogonal is a line which connects places with the same magnetic variation, the agonic line is the line of zero magnetic dip
An isogonal is a line which connects places with the same magnetic variation, the aclinic is the line of zero magnetic dip
35.
Given: True HDG = 035°,TAS = 245 kt,Track (T) = 046°,GS = 220 kt.Calculate the W/V?
335/45kt
340/45kt
335/55kt
340/50kt
36.
Isogonals are lines of equal
pressure.
wind velocity
compass deviation
magnetic variation
37.
The nominal scale of a Lambert conformal conic chart is the
mean scale between the parallels of the secant cone
scale at the equator
mean scale between pole and equator
scale at the standard parallels
38.
Given:FL120, OAT is ISA standard, CAS is 200 kt,Track is 222°(M),Heading is 215° (M),Variation is 15°W.Time to fly 105 NM is 21 MIN.What is the W/V?
055°(T) / 105 kt
040°(T) / 105 kt.
065°(T) / 70 kt.
050°(T) / 70 kt
39.
Given: TAS = 198 kt,HDG (°T) = 180,W/V = 359/25.Calculate the Track(°T) and GS?
180 - 223 kt
180 - 183 kt
179 - 220 kt
181 - 180 kt
40.
61.4.7.0 (4483) An aircraft was over 'A' at 1435 hours flying direct to 'B'.Given:Distance 'A' to 'B' 2900 NMTrue airspeed 470 ktMean wind component 'out' +55 ktMean wind component 'back' -75 ktSafe endurance 9 HR 30 MIN The distance from 'A' to the Point of Safe Return (PSR) 'A' is
2141 NM
1611 NM
2844 NM
1759 NM
41.
An aircraft takes-off from an airport 2 hours before sunset. The pilot flies a track of 090°(T), W/V 130°/ 20 kt, TAS 100 kt. In order to return to the point of departure before sunset, the furthest distance which may be travelled is
105 NM
115 NM
84 NM
97 NM
42.
The main reason for mounting the detector unit of a remote reading compass in the wingtip of an aeroplane is
to ensure that the unit is in the most accessible position on the aircraft for ease of maintenance
to minimise the amount of deviation caused by aircraft magnetism and electrical circuits
to maximise the units exposure to the earth's magnetic field
by having detector units on both wingtips, to cancel out the deviation effects caused by the aircraft strucure
43.
The flight log gives the following data :""True track, Drift, True heading, Magnetic variation, Magnetic heading, Compass deviation, Compass heading""The right solution, in the same order, is
117°, 4°L, 121°, 1°E, 122°, -3°, 119°
115°, 5°R, 120°, 3°W, 123°, +2°, 121°
119°, 3°L, 122°, 2°E, 120°, +4°, 116°
125°, 2°R, 123°, 2°W, 121°, -4°, 117°
44.
An aircraft is flying with the aid of an inertial navigation system (INS) connected to the autopilot. The following two points have been entered in the INS computer:WPT 1: 60°N 030°WWPT 2: 60°N 020°WWhen 025°W is passed the latitude shown on the display unit of the inertial navigation system will be
59°49.0'N
60°00.0'N
60°11.0'N
60°05.7'N
45.
Given:Magnetic heading 311°Drift angle 10° leftRelative bearing of NDB 270°What is the magnetic bearing of the NDB measured from the aircraft?
208°
221°
180°
211°
46.
What is the final position after the following rhumb line tracks and distances have been followed from position 60°00'N 030°00'W?South for 3600 NM, East for 3600 NM, North for 3600 NM, West for 3600 NM.The final position of the aircraft is
60°00'N 090°00'W
59°00'N 090°00'W
59°00'N 060°00'W
60°00'N 030°00'E
47.
On a polar stereographic projection chart showing the South Pole, a straight line joins position A (70°S 065°E) to position B (70°S 025°W).The true course on departure from position A is approximately
135°
225°
315°
250°
48.
An aircraft departs from position A (04°10' S 178°22'W) and flies northward following the meridian for 2950 NM. It then flies westward along the parallel of latitude for 382 NM to position B. The coordinates of position B are?
53°20'N 172°38'E
53°20'N 169°22W
45°00'N 172°38'E
45°00'N 169°22W
49.
In an Inertial Navigation System (INS), Ground Speed (GS) is calculated:
from TAS and W/V from Air Data Computer (ADC)
by integrating gyro precession in N/S and E/W directions respectively
from TAS and W/V from RNAV data
by integrating measured acceleration
50.
Given:Runway direction 305°(M),Surface W/V 260°(M)/30 kt. Calculate the crosswind component?
18 kt
24 kt
27 kt
21 kt
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