Ground School Exam Questions & Question Banks

[atpcliff]

Hi!

Thanx everyone 4 posting! Some of the questions are helpful, as is the knowledge that I'm studying the correct stuff.

I have 5 days to study and pass ALL the subject materials condensed into one Conversion Exam, NOT through my own doing!!! I would've been studying months ago if I had the resources!

So, I'm reading the questions and trying to learn everything I don't know via the questions and answers. Not very nice! But, luckily, I went to university, so I can give it the "Old College Try"!

It IS frustrating having to learn stuff that doesn't apply to me at all, like recip engines and polar charts, but, too bad for me! I want to pass, so I have to learn it all.

cliff
NBO
PS-Saw the one poster who is VC10. Just saw one of those yesterday, and we had to ask what it was...could not remember!

[atpcliff]

Hi!

Just changed my exam to tommorrow, so 3 days to study for the entire knowledge base

Extra Nice!

cliff
NBO

[atpcliff]

Hi!

Just passed...WHEW!!!...the Conversion Exam.

Briefly:
Studied for Air Law exam to get License Validation back in Spring. Supposed to take Air Law, but scheduled for Conversion Exam (covers ALL subjects...Air Law, ALL PPL, ALL CPL, ALL ATPL exam material) by mistake. Took Conversion Exam, and scored 59. Had ONLY studied for Air Law.

Passed Air Law a couple of weeks later.

This past week told taking Conversion Exam, and then was given today's date.

So, Studied about 3 days, and took the exam today: ALL the subjects, and passed. Inshallah!!!

It used to be that all you had to do to convert your license was pass the Air Law exam, but this year, they changed it, so need the full Conversion Exam to convert your license. Passing the Air Law is a requirement now to get a temporary Validation of your home-country license. I believe most countries still allow you to convert with only the Air Law.

So glad I passed!!!...Thanx a TON to everyone that posted info, questions, and answers on this thread...it helped a lot.

Odd question:
How old to get a CPL Examiner's License...I had no idea...guessed age 25.

cliff
NBO
PS-Thanks AGAIN!!!

[filmarik]

Hi folks, could anybody help me please?

The operator of an aircraft equipped with 50 seats uses standard masses for passengers and baggage. During the preparation of a scheduled flight a group of passengers present themselves at the check-in desk, it is apparent that even the lightest of these exceeds the value of the declared standard mass.

A the operator may use the standard masses for the balance but must correct these for the load calculation

B the operator may use the standard masses for the load and balance calculation without correction

C the operator should use the individual masses of the passengers or alter the standard masss

D the operator is obliged to use the actual masses of each passenger


AND

The maximum zero-fuel mass:
1- is a regulatory limitation
2- is calculated for a maximum load factor of +3.5 g
3- is due to the maximum permissible bending moment
at the wing root
4- imposes fuel dumping from the outer wings tank first
5- imposes fuel dumping from the inner wings tank first
6- can be increased by stiffening the wing
The combination of correct statements is:

A

1, 2, 3

B

1, 3, 5

C

2, 5, 6

D 4, 2, 6


Thanks a lot

[filmarik]

Hi folks, could anybody help me please?

The operator of an aircraft equipped with 50 seats uses standard masses for passengers and baggage. During the preparation of a scheduled flight a group of passengers present themselves at the check-in desk, it is apparent that even the lightest of these exceeds the value of the declared standard mass.

A the operator may use the standard masses for the balance but must correct these for the load calculation

B the operator may use the standard masses for the load and balance calculation without correction

C the operator should use the individual masses of the passengers or alter the standard masss

D the operator is obliged to use the actual masses of each passenger


AND

The maximum zero-fuel mass:
1- is a regulatory limitation
2- is calculated for a maximum load factor of +3.5 g
3- is due to the maximum permissible bending moment
at the wing root
4- imposes fuel dumping from the outer wings tank first
5- imposes fuel dumping from the inner wings tank first
6- can be increased by stiffening the wing
The combination of correct statements is:

A
1, 2, 3
B
1, 3, 5
C
2, 5, 6

D 4, 2, 6


Thanks a lot

[tumboseppala]

Hi there. Sorry to bring old topic up but for filmarik:

For the first one I would go with D.

Number 2 would be B.
MZFM is a regulatory limit so 1 must be correct.
The maximum DLL for ac is +2,5g, so +3,5 is out of limits.
3 is correct since bending moments are greatest at wing root. Adding mass like engines and fuel, decrease these moments so 3 is correct. From there you can work it out if 1&3 is correct then 5 must be correct and that is also true. Fuel is used from center towards the wingtip in all conditions.

Please correct if I'm wrong.

[shivamjoshi]

Can anyone please help me with questions involving calculation of altitudes at different pressure levels?

What would be the true altitude at 100hpa and 200hpa at standard temp.?

I know the lapse rate from 0 to 10,000ft. is 27ft./1hpa and from 10,000ft.-20,000ft. is 47ft./1hpa. Don't Remember the lapse rates after 20k.

Thanks

[Mzehmed]

Principles Of Flight Exam Question.

Hi guys, just a quick P.O.F question from me.

Question:

A 50 ton twin-engine aeroplane performs a straight, steady, wings level climb. If the lift/drag ratio is 12 and the thrust is 60000N per engine, the climb gradient is: (assume g=10m/sē).

I believe I have the correct formulae to work with as:

Gradient Of Climb = Thrust - Drag / Weight (x100)

Therefore we have;

Thrust = 120,000N (60,000x2)
Weight = 50,000 KG

Gradient Of Climb = 120,000 - (Drag)* / 50,000 (x100)

*At this point I am fairly lost and not sure how to calculate the drag to enter into the formulae. Just as extra info the options for this answer are:

a) 3.7%
b) 15.7%
c) 12%
d) 24%

Thanks for any advice.

[tigermagicjohn]

You have the formula correct, however you need to know the Lift/Drag ratio calculation.

This is 1/12 x 100, which will give you - the number 8.333
Formula is correct, but can be confusing to do the calculation. The lift/drag ratio will always be as above, example 1/8 x 100 for 8 lift drag ratio.

What you do which I found better is:

(120.000N / 500.000)( x 100)) - 1/12 x 100 = Climb gradient

(0.24) x 100 - (0.083 x 100) - 15.67%

Better to write:

Climb gradient = Thrust / Weight x 100% - (Lift / Drag) x100%

Weight is mass x gravity, 50.000 kg x 10 (not 100)

so the first part is actually:
(Thrust / (Mass x Gravity)) (x 100%)

Remember Weight is not 50.000 kg, Weight is 500.000N in this example.
Mass and Weight are not the same.

[Keith.Williams.]

The correct equation is

% Climb gradient = approximately 100% ( (Thrust - Drag)/Weight)

Lift is less than weight in a steady climb, but for small climb angles the difference is very small. If we ignore this small difference we can say that lift = weight.

We can now resate the above equation as

% Climb Gradient = 100% ((Thrust/Weight) - (Drag/Lift))

But Drag/Lift is just 1 / Lift to Drag ratio

So we can restate the equation as

% Climb Gradient = 100% ((Thrust/Weight) - (1/ Lift to Drag ratio))

In this question we get

% Climb Gradient = 100% ((120000 N / 500000 N) - (1/12)

% Climb Gradient = 15.67%

All of the above is of course an approximation based on the simplfying assumption that lift = weight.

But it gives us a nice easy equation which bypasses the need to calculate the drag.

% Climb Gradient = 100% ((Thrust/Weight) - (1/Lift to Drag Ratio))

[aris214]

I am also having difficulty with this question. Any help would be appreciated.

"What would be the true altitude at 100hpa and 200hpa at standard temp.?"

[Keith.Williams.]

ISA table gives 38661 ft for 200 hPa.

If you do a google search for ISA you will find various computation programs. One I tried gives 68400 ft for 100 hPa.