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CPL Test Series
Question Bank
Questions for DR ELEMENT
Answer the following questions
Test Mode
Training Mode
1. The pressure altitude of the field can be found by?
Setting 1013 mb on the altimerter subscale.
Setting QFE on the altimeter subscale.
Setting QNH on the altimeter subscale.
2.
An A/C takes off from A with altimeter reading 240'. Elevation A = 330'. QFEA = 997 hPa. It goes to B. Elevation B = 300'. Altimeter reads 210 at landing, without changing the setting. Find QNH and QFE at B.
- [INDICATED = 1005 HPA QFEB = 998 HPA, READS = 210'.]
- [INDICATED = 1005 HPA QFEB = 998 HPA, READS = 211'.]
3. If pressure increases whilst temperature increases in the ISA
Density will remain constant.
Density might Increase or decrease.
Density will increase.
4. If pressure altitude is 0 ft. and ambient temperature is 10°C, what is the density altitude?
593 ft.
1593 ft.
-593ft.
5.
On 3° GP at DME, ILS glide slope show 2 dot fly up indication on 4 dot indication. Approximately how many feet a/c is above/below G/S (glide slope?
- [H = D × GA × 101, GA = 0.35°, D = 3NM, H = 0.35× 3× 101 = 109' BELOW GP]
- [H = D × GA × 101, GA = 0.35°, D = 3NM, H = 0.35× 3× 101 = 106' BELOW GP]
6. A/C is flying inddicated altitude of 21,000' OAT = -12°C. Find TA
- [ISA = - 42°+15°=-27°, TA = 21,000' + [4×(12° - (27°))× 21,000'/ 1,000'], = 21,000' + [4×15×21], = 21,000'+1,260' = 22,360'.]
- [ISA = - 42°+15°=-27°, TA = 21,000' + [4×(12° - (27°))× 21,000'/ 1,000'], = 21,000' + [4×15×21], = 21,000'+1,260' = 22,260'.]
7.
At 4.8 nm ILS/DME, G/S shows 2 dot fly up indication on 5 dot indicator. There is an obstruction of elevation of 450'. A/C altimeter is set to QNH of the RWY. RWY elevation is 300ft. Find (a) Clearance of a/c while passing over obstruction. (b) Altimeter reading over obstruction. (c) Ht. of a/c over obstruction.
- [D = 4.8NM, GA = 3-0.28 = 2.72°, H = 4.8×2.72×101 = 1319'. ALTITUDE = 1319' + 300' = 1619'. A) CLEARANCE = 1619' - 450' = 1169'. B) ALTIMETER = 1619', HEIGHT = 1319'. ]
- [D = 4.8NM, GA = 3-0.28 = 2.72°, H = 4.8×2.72×101 = 1319'. ALTITUDE = 1319' + 300' = 1619'. A) CLEARANCE = 1619' - 450' = 1169'. B) ALTIMETER = 1619', HEIGHT = 1419'. ]
8. Density altitude is?
The elevation at which the existing density would occur in the ISA.
The altitude at which the existing density would occur in the ISA.
The density at which the existing temperature would occur in the ISA.
9.
An A/C takes off from place A. QNHA = 998 hPa elevation A = 90'. Altimeter Reads (-) 90' at the time of take off. It goes to place B. QNB = 1020 hPa elevation B = 200'. Mid way there is a hill of elevation 3000 mtrs. Find: (a) Subscale reading at take off. (b) QNH at hill (assume uniform pressure gradient). (c) When indications of altimeter will give clearance of 500' while crossing over hill.
- [INDICATED = 992 ANHHILL = 1009, DIFF = 17 × 30 = 510' READING = 9840' + 500' = 10,340', READING FOR CLEARANCE = 10,340 - 510 = 9,831'.]
- [INDICATED = 992 ANHHILL = 1009, DIFF = 17 × 30 = 510' READING = 9840' + 500' = 10,340', READING FOR CLEARANCE = 10,340 - 510 = 9,830'.]
10. If QFE is 1022 hPa what is the pressure altitude of the filed?
30660 ft amsl
-270 ft amsl
270 ft amsl
11. If QNH is 999 hPa, what is the pressure altitude at an elevation of 25000 ft.
25300 ft.
25200 ft.
25400 ft.
12. Density varies:
Inversely with pressure and directly with temperature
Directly with pressure and inversely with temperature
Inversely with temperature and pressure
13. Take-off and landing performance must be calculated based on ____ and ____ respectively?
Actual temperatures Actual temperatures
Forecast temperatures Forecast temperatures
Actual temperatures Forecast temperatures
14. Pressure altitude 29000 ft, OAT -55°C. calculate the density altitude ?
27500 ft.
28900 ft.
27850 ft.
15.
An a/c takes off from place A with altimeter reading 180'. Elevation of A is 90'. QNH of A 1020 hPa, a/c climbs to altimeter indications of 8000'. A/C goes to B. Elevation B = 30'. QFEB = 998 hPa. (a) Subscale setting at A (b) True altitude of a/c over B. (c) True height of a/c over B. (d) QNHB (e) Altimeter reads 120' while landing at B. Find altitude with altimeter subscale set at 999'hPa at B.
- [(A) SUBSCALE = 1023 HPA. (B) TRUE ALT. = 8,000' - 720' = 7,280'. (C) HEIGHT = 7,280' - 30' = 7,250' (D) QNHB = 999 HPA. (E) ALTITUDE = - 600']
- [(A) SUBSCALE = 1023 HPA. (B) TRUE ALT. = 8,000' - 720' = 7,280'. (C) HEIGHT = 7,280' - 30' = 7,250' (D) QNHB = 999 HPA. (E) ALTITUDE = - 610']
16. In atmosphere rate of fall of pressure is
Exponential
Linear
Progressive
17. If the OAT at a pressure altitude of 5000 ft. amsl is 10°C what is the temperature deviation?
-4.90C.
+4.90C.
-14.90C
18. The atmospheric pressure at FL70 in a "standard + 10" atmosphere is
1013.25hPa
781.85hPa
942.13hPa
19.
At 2.7 DME G/S shows 1/2 of full scale fly down indications. At this place there is a spot height of 200ft. Find clearance of a/c while flying over spot height.
- [GA = 3° + 0.35° = 3.35° D = 2.7 NM. H = 2.7×3.35×101 = 914'. CLEARANCE = 914 - 200 = 734'. ]
- [GA = 3° + 0.35° = 3.35° D = 2.7 NM. H = 2.7×3.35×101 = 914'. CLEARANCE = 914 - 200 = 714'. ]
20. If aerodrome elevation is 4000 ft. amsl, and QNH is 1025 hPa, what is the pressure altitude?
3640 ft.
3740 ft.
3540 ft.
21. In ISA conditions, with increase in altitude rate of fall of pressure
Decreases
Remains same
Increases
22. An a/c is flying indicated altitude of 25,000 ft. SAT = - 50°C. Find true altitude
- [ISA TEMP AT 25,000' = 25°× 2°=-50°+15°=-35°. DEV = -50° - (-35°) = - 15°. = 25,000' + [4×(-15°)×25,000 / 1,000] = 25,000'-1,500'. = 23,500'.]
- [ISA TEMP AT 25,000' = 25°× 2°=-50°+15°=-35°. DEV = -50° - (-35°) = - 15°. = 25,000' + [4×(-15°)×25,000 / 1,000] = 25,000'-1,500'. = 23,600'.]
23. If field elevation is 1000 ft. amsl, QNH is 1025 mb, an QFE is 991.67 mb, what is the pressure altitude at the field?
-640 ft.
1640 ft.
640 ft.
24. A/C is flying at PA = 17,000', Ambient Temp = -30°C. Find TA.
- [ISA TEMP = -34°+15°= - 19°, =TA = 17,000'+[-4 ×11 × 17], =17,000' - 748' = 16,252'.]
- [ISA TEMP = -34°+15°= - 19°, =TA = 17,000'+[-4 ×11 × 17], =17,000' - 748' = 16,242'.]
25. Which of the following cause air density to decrease?
Increasing humidity, decreasing altitude, increasing temperature.
Increasing humidity, increasing altitude, increasing temperature.
Decreasing humidity, increasing altitude, decreasing temperature.
26.
An a/c takes off from place A with elevation 90' with altimeter reading 0'. QNHA = 1020 hPa. It climbs to 8000'. It goes to B. Elevation B = 30'. QNHB = 1004 hPa. Find (a) True altitude of A/C over B (b) Height of a/c over B (c) Altimeter reading while landing at B. (Subscale was not reset)
- [(A) SET 1017. TA OVER B = 7,610' (B) HEIGHT OVER B = 8000' -420 = 7,580' (C) ALTIMETER WHILE LANDING AT B = 410']
- [(A) SET 1017. TA OVER B = 7,610' (B) HEIGHT OVER B = 8000' -420 = 7,580' (C) ALTIMETER WHILE LANDING AT B = 420']
27. What is the true altitude if OAT = +35° C. and pressure altitude =5000 feet ?
4555 ft.
5320 ft.
5550 ft.
28. If the temperature is 12°C at a pressure altitude of 10000 ft, what is the density altitude?
14993 ft.
13993 ft
11993 ft.
29. The ISA temperature and pressure for 18,000 ft are
-20.7° C and 9=595.2 hPa respectively
-12.7° C and 506 hPa respectively
-20.7° C and 506 hPa respectively
30. If pressure altitude is 37000 ft amsl and QNH is 1000 mb, what is field elevation?
490.
3310.
390.
31. If QNH changes from 1013 hPa to 1022 hPa will?
Decrease field elevation.
Not affect field elevation.
Increase field elevation.
32. What is the ISA temperature value at FL 330?
-50°C.
-53°C.
-56.5°C.
33. An aircraft flies an altitude of 3500 feet from A to B, elevation 700 feet QNH 1015 hPa. B Elevation is 1120 feet QNH 992 hPa. Assuming the altimeter sub-scale is not changed, the aircraft will arrive over B at a height of
1690 feet
670 feet
2810 feet
34.
At 3nm ILS/ DME localiser shows 2 dot fly right indications and G/S shows 2 dot fly up indication on 5 dot indicator. A/C ground speed is 180 kmph. Find (i) How many mtr a/c is left right and above / below glide slope. (ii) Rate of descend of a/c (ROD)
- [D = 3NM, GA = 0.28, (I) H = 3×028×101 = 85' OR 85/3.28M = 26M BELOW GP. DISPLACEMENT FROM CENTER LINE = 3×1×101 = 303' = 303/3.28M = 92M LEFT, (II) ROD = 180×0.54×3×1.69 = 492' PER MIN.]
- [D = 3NM, GA = 0.28, (I) H = 3×028×101 = 85' OR 85/3.28M = 26M BELOW GP. DISPLACEMENT FROM CENTER LINE = 3×1×101 = 303' = 303/3.28M = 92M LEFT, (II) ROD = 180×0.54×3×1.69 = 494' PER MIN.]
35.
An A/C takes off from A with altimeter reading 240'. A elevation 300' QFEA = 994 hPa, goes to B. Elevation B = 180'. Altimeter reads 150' at landing, with out changing the setting. Find QNH at A and QNH and QFE at B.
- [INDICATED : 1002 INDICATED : 1002 HPA, QNHB = 1003 → 180', QFEB = 997 HPA]
- [INDICATED : 1002 INDICATED : 1002 HPA, QNHB = 1003 → 180', QFEB = 999 HPA]
36.
A/C takes off from A elevation A = 90'. PA of A = 150' It goes to B, PAB = 450'. Elevation B = 240'. Find QNH & QFE at A and B.
- [QNHA 1011 → 90'. QFEA = 1008'. QNHB=1006 → 240' QFEB = 898.]
- [QNHA 1011 → 90'. QFEA = 1008'. QNHB=1006 → 240' QFEB = 998.]
37. An aircraft is in level flight at Fl 100 over a mountain range, which extends up to 2400 meters AMSL. If the regional QNH is 998 hPa (use 30 ft/hPa), what is the approximate terrain clearance?
450 feet.
7869 feet.
1681 feet.
38. If field pressure altitude is 5000 ft amsl and OAT is 25°C. What is the density altitude?
5000 + 118 - (25 - ( 15 - (5 × 1.98))) = 7348.2 ft.
5000 + 118 - (25 + ( 15 - (5 × 1.98))) = 8551.8 ft.
5000 + 118 - (25 - ( 15 + (5 × 1.98))) = 4988.2 ft.
39.
Distance between A & B 1000 nm. QFEA = 990 h Pa elevation A = 90'. A/C takes off with QNH setting at A. 700 nm from A, there is a hill a 7,000' elevation. QFEB 1010 hPa elevation B = 600'. Find (a) What indicated altitude will give clearance of 1,000' over hill. (assume uniform pressure gradient) (b) What will altimeter show, while landing at B (altimeter was not reset)
- [QNHHILL = DIFF 37/1000 × 700 = 1019 HPA. 7,000+1,000=8,000 SETTING = 993 HPA. DIFF 30 × 37 = 1110', INDICATION (-) 590'.]
- [QNHHILL = DIFF 37/1000 × 700 = 1019 HPA. 7,000+1,000=8,000 SETTING = 993 HPA. DIFF 30 × 37 = 1110', INDICATION (-) 510'.]
40. If pressure altitude is 22,800 ft, at an elevation of 22,000 ft, what is QNH?
1034 hPa.
985 hPa.
1026 hPa.
41.
At 3.6 nm ILS/DME. G/S shows 2 dot fly up indication on 5 dot indicator. A/C altimeter is set to QNH of threshold. RWY elevation 300', a/c ground speed = 200 kmph. Find (i) Altimeter reading at present position (ii) ROD
- [D = 3.6NM, GA = 3.028 = 2.72°, H = 3.6 × 2.72 × 101 = 989' (I) ALTITUDE = 989' + 300' = 1289'. (II) ROD = 200×0.54×3×1.69 = 548' PER MIN]
- [D = 3.6NM, GA = 3.028 = 2.72°, H = 3.6 × 2.72 × 101 = 989' (I) ALTITUDE = 989' + 300' = 1289'. (II) ROD = 200×0.54×3×1.69 = 549' PER MIN]
42. Density altitude is? 1. The elevation at which the prevailing density occurs in the ISA 2. The Pressure altitude at which the prevailing temperature occurs in the ISA. 3. The Pressure altitude at which the prevailing density occurs in the ISA.
1
1 and 3.
1 and 2.
43. If pressure increases whilst temperature increases in a non-standard atmosphere
Density might increase decrease.
Density will decrease.
Density will increase.
44. If QFE changes from 1013 hPa to 1022 hPa will?
Not affect QNH.
Increase field elevation
Increase QNH.
45.
An a/c takes off from place A with altimeter reading 0'. QNHA = 1015 hPa. Climbs to altimeter indications of 6,300' and goes to place B. QNHB = 1006 hPa elevation at A is 60' and elevation of B is 300', Find (a) Subscale reading while taking off from A. (b) True altitude of a/c over B. (c) True ht. of a/c over B. (d) Indicated Altitude while landing at B. (subscale not reset).
- [(A) 1013 HPA (B) TRUE ALT = 6,300 - 210' = 6,090'. (C) HT = 6,090' - 300' = 5,790' (D) ALTIMETER AT B = 300' + 210' = 510.]
- [(A) 1013 HPA (B) TRUE ALT = 6,300 - 210' = 6,090'. (C) HT = 6,090' - 300' = 5,790' (D) ALTIMETER AT B = 300' + 210' = 590.]
46. As altitude increase in the ISA?
The effects of decreasing density outweigh those of decreasing pressure and temperature so aircraft performance decreases
The effects of decreasing pressure outweighs those of decreasing temperature, so aircraft performance decreases.
The effects of increasing TAS: CAS ratio outweigh those of pressure and temperature so aircraft performance decreses.
47. On board an aircraft the altitude is measured from the
Pressure altitude
Standard altitude
Density altitude
48. An a/c is flying with indicated altitude 20,000 ft, OAT = -10°C. Find True Altitude
- [ISA TEMP AT 20,000' =(20°×2°) = -40°+15° = -25°. DEV = -10° - (-25°) = 15°. TRUE ALT = PA + [4×(DEV)×PA/1000], = 20,000 + [4×15×20,000/1000], = 20,000' + (1200') = 21,200'. THUMB RULE : TRUE ALTITUDE VARIATION IS 4% OF PRESSURE ALTITUDE FOR EVERY 10°C DEVIATION FROM ISA.]
- [ISA TEMP AT 20,000' =(20°×2°) = -40°+15° = -25°. DEV = -10° - (-25°) = 15°. TRUE ALT = PA + [4×(DEV)×PA/1000], = 20,000 + [4×15×20,000/1000], = 20,000' + (1200') = 21,200'. THUMB RULE : TRUE ALTITUDE VARIATION IS 4% OF PRESSURE ALTITUDE FOR EVERY 12°C DEVIATION FROM ISA.]
49.
An a/c takes off from place X with altimeter set at 1008 hPa. And climbs to indicated altitude of 9,500' it goes to place Y where QNH is 1015 hPa. Elevation of Y is 400'. Find: (a) True altitude of A/C over Y. (b) Height of a/c over Y. (c) Altimeter reading on landing at Y (subscale was nor reset to QNH of Y).
- [(A) TRUE ALTITUDE WILL BE HIGH 9,500' + 210' APPX = 9710 (B) HEIGHT = 9,710' - 400' = 9,310'. (C) 400' - 210' = 191'.]
- [(A) TRUE ALTITUDE WILL BE HIGH 9,500' + 210' APPX = 9710 (B) HEIGHT = 9,710' - 400' = 9,310'. (C) 400' - 210' = 190'.]
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