guyz can u plz help me these ques
 

An aircraft is flying at FL 140 where the COAT is -5° C. It is flying at an indicated air speed of 260 kts and is experiencing a headwind of 34 knots. When 150 nm from the FIR boundary it is instructed to reduce speed in order to delay arrival at the boundary by 5 minutes. The required reduction in indicated air speed is:
A) 33 kts.
B) 41 kts.
C) 24 kts.
D) 15 kts.
ans a
The time interval between the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the aircraft is 2 milli seconds. The slant range is:
A) 162 nm
B) 73 nm
C) 323 nm
D) 92 nm

ans is 162nm

IAS or CAS = 260 kts Find TAS which = 323 kts wind = 34 Headwind therefore Ground speed = 289 kts , Now time taken to cover 150Nm @ 289 = 31min.wit a delay of 5 min revised time = 36 min . Now speed = Distance/ Time therefore Speed @ delay of 5min= 150/36 min = 250 kts (ground speed) now convert 250 kts to TAS by adding 34 kts headwind which give you TAS of 284 kts using your flight computer Find CAS @ that TAS which equates to 228 kts ,,,,

now find the difference : 260 Kts -228Kts = 32 kts / 33kts :rolleyes:

An aircraft is flying at FL 140 where the COAT is -5° C. It is flying at an indicated air speed of 260 kts and is experiencing a headwind of 34 knots. When 150 nm from the FIR boundary it is instructed to reduce speed in order to delay arrival at the boundary by 5 minutes. The required reduction in indicated air speed is:
A) 33 kts.
B) 41 kts.
C) 24 kts.
D) 15 kts.
ans a
The time interval between the transmission of a given DME interrogation pulse and the reception of the appropriate response pulse at the aircraft is 2 milli seconds. The slant range is:
A) 162 nm
B) 73 nm
C) 323 nm
D) 92 nm

ans is 162nm


the first one seems dodgy but colsest i came was using cx2 using actual tas and a CAS of 226 ...i got an ans of 36 kts difference

2nd one is simple

distance = speed/time = 3*10^8 / 0.002 = 324
but this equals distance wave travels to stn and back so we divide by 2 we get 162

Maverick2167
 

thanks for the help bro

Shabez
 

thanks for the help

CONTACT NUMBER FOR CAE PEOPLE
 

Hii friends i have been trying to get a hall number for indigo assessment for pretty long time now ..can sum1 plz PM the contacts number for the cae people

any info will be appreciated

Where are all the above questions found exactly??

just received my hall ticket e-mail from Capt Upendranath for the 20th Oct..

Butterfly whats the venue this time..

same as earlier..that "parkland hotel"..wish they'd found a better place!!

Thanks buddy!

edit: Just got my hall ticket..for 21st

help needed

Whats the formula to calculate NDB range??

The nominal maximum range of an NDB with a transmitter power is 200 watts is:

A) 150 to 170 nm.
B) 200 to 220 nm.
C) 100 to 120 nm.
D) 50 to 60 nm.

the formula for NDB range is
over water (nm) = 3√Wattage
Over Land (nm) = 2√Wattage

So according to me the range of a 200 W NDB is 2√200 =28.3nm

is the answer in Km or nm??

nms i suppose!!
none of the ans match with these formulae?!?!?!?
another Q

The approximate range of a 10 KW NDB over the sea is:

A)100 nm

It must be 1KW
Range over water = 3√1000 =94.8 =approx 100nm
where are you getting these questions from?

aviationtire..where r u studying from??

though many of their qs do contradict each other!!

Just the books.
i dont think all their answers are correct

help please..

An aircraft is over flying a VOR at 30.000 ft, at a groundspeed of 300 kt. The maximum time during which no usable signals will be received (in minutes and seconds) is:

A)0.50
B)4.40
C)2.25
D)1:40

guys can u plz help
 

A cambered airfoil with zero angle of attack, will in flight produce:
A) some lift and some drag.
B) no lift and no drag
C) some lift with no drag.
D) no lift but some drag.

@ buttefly747
 

it has been answerd 4 days ago

http://www.pprune.org/south-asia-far...questions.html

@flying machine
 

The answer is a.

A cambered aerofoil produces lift even at 0 angles of attack. On a typical cambered aerofoil, -4 degs produces 0 lift. And since there is lift at 0 degrees, there is induced drag too.

Quick question for those who have applied for the program and got a call for the exam. I applied on the 21st of September for the Indigo-Cae program but did not get a call for this exam.

How long after you guys applied did you get the first call for the exam?

I got a call for the exam on the 20th:D

@superdrunker
 

it really depends. coz its just a computer generated call

vmendes
 

thanks mate

Just trying to figure out what the average is.

guys can u plz help
 

On a Lambert conformal chart the distance between two parallels of latitude having a difference of latitude = 2° , is measured to be 112 millimetres. The distance between two meridians, spaced 2° longitude, is, according to the chart 70 NM. What is the scale of the chart, in the middle of the square described?
A) 1 : 1 056 000
B) 1 : 756 000
C) 1 : 1 984 000
D) 1 : 1 233 000
ans c
how exactly thats has come could you please tell me

Flying machine
 

Why don't you put up your questions in the 'General Navigation questions' thread ? In that way, it is much easier for others to track all the questions as well. :ok:

bieng human
 

i am sorry but i am not aware of that thread can please tell me the link aswell
thanks

Hall Ticket?
 

Hey I got second mail on 14 oct. but didnt get hall ticket:(
Any one else in the same situation waiting for final mail:ugh:
Please reply
ala

Help!!
 

A pilot receives the following signals from a VOR DME station: radial 180° +/- 1° , distance = 200 NM. What is the approximate error?
A) +/- 3.5 NM. [correct answer]

B) +/- 7 NM.

C) +/- 2 NM.

D) +/- 1 NM.

mine comes out to be 2.5nm

Use 1 in 60 rule and use degree as 1

200/60 = 3.33 ~ 3.5

has anyone received the hall ticket for 21st oct apart from DJ flyboy???

climb ... due to the inversion which means warmer air is present above.

I think it is 150..since there is no wind mentioned we cant factor it in.

To the guys who did not get a call do not worry there will be atleast one more exam if not two before the end of this year.

They had one A/C delivered on the 11th of this month and still 3 more to come by end of this year.

So keep studying hard and be patient!
:)

@liedetector
 

Indigo is gonna conduct exam every month for sure. Next assessment is on nov 3rd week. dates not fixed yet

hey guys, all the best for tomorrow.:)

Please help solve this question...
The center of a small island is identified at the intersection of 60 degree left bearing line and 15nm range arc of airborne weather radar. If aircraft's heading and height are 035 degrees (M) and 42500 feet. What QTE and range should be plotted in order to obtain a fix from the island? Var - 20W.

Ans - 135 degrees and 13nm.

035(m)-60 =335 (m)
Var = 20w therefore True bearing is 315
QTE = 315-180=135

42500feet = 7nm
Use Pythag Distance = sqrt(15x15 - 7 x 7) =13.26 nm