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The correct option is (a) True

To EXPLAIN I would SAY: The TERMINAL phases CONSIST of two parts. They are ground run distance and AIRBORNE distance. The terminal phases of the aircraft are the take-off phase at the departure and landing at the destination. In the take-off phase the aircraft is transformed to safe airborne state from the stationary ground state.

Correct OPTION is (c) Take-off and landing

To ELABORATE: The TERMINAL phases of the aircraft are the take-off phase at the DEPARTURE and landing at the destination. In the take-off phase the aircraft is transformed to safe AIRBORNE state from the stationary ground state.

Correct answer is (a) It is the ratio of SQUARE of wing SPAN to wing area

Best explanation: Aspect ratio is the ratio of the square of wing span to wing area. It is given by the formula, AS=\(\frac{b^2}{S}\) where AS is aspect ratio, b is wing span and S is wing area.In general the aspect ratio is maintained high in the AIRCRAFT for structural RIGIDITY and REDUCING drag.

The CORRECT choice is (a) military aircrafts

The BEST explanation: Ski-jump take-off path is the path which has an elevation at the end of the runway i.e. the ROAD is lifted up with respect to GROUND at an angle. This ski-jump assisted angle is approximately EQUAL to 12°.

Right answer is (a) 5

To elaborate: The ASPECT RATIO is given by the ratio of span and chord. Given span = 10 feet and chord is 2 feet. SUBSTITUTING in the formula we get AS=\(\frac{span}{chord}\). AS=\(\frac{10}{2}\).

AS=5.

The correct answer is (a) REDUCED take-off and LANDING, short take-off and landing, vertical take-off and landing

Explanation: There are five types of take-off and landings. They are:

CTOL- convectional take-off and landing

RTOL- reduced take-off and landing

STOL- short take-off and landing

VTOL- vertical take-off and landing

STOVL- short take-off and vertical landing.

Correct choice is (a) True

The best I can explain: The AIRCRAFT uses short take-off landing to reduce the take-off FUEL DEMAND. This results in take-off of the aircraft with the WEIGHT of the aircraft above maximum VERTICAL take-off and landing (VTOL) weight.

The correct option is (a) True

The explanation is: High lift devices can be USED to increase the maximum lift coefficient of an AIRCRAFT and reduce the lift off SPEED. The heavy lift devices INCLUDE full span flaps, drooped ailerons and passive BOUNDARY layer control systems.

The CORRECT choice is (a) True

The best explanation: VTOL is the EXTREME case of STOL. The AIRCRAFT uses short take-off landing to REDUCE the take-off fuel demand. This results in take-off of the aircraft with the weight of the aircraft above maximum VERTICAL take-off and landing (VTOL) weight.

The correct option is (d) elevator

The explanation: High lift devices can be used to increase the maximum lift COEFFICIENT of an AIRCRAFT and reduce the lift off speed. The heavy lift devices include full SPAN flaps, drooped ailerons and PASSIVE boundary LAYER control systems.

Correct CHOICE is (b) installing the propeller in front of the wing and deflecting the slip STREAM with large FLAPS

The explanation: The power- induced LIFT is generated by installing the propeller in front of the wing and deflecting the slip stream with large flaps. This way helps in wing area in the propeller SLIPSTREAM to generate considerable lift.

The correct ANSWER is (a) The airspeed below which the rudder the will not be capable of producing a YAWING MOMENT

To explain I would say: The minimum CONTROL speed ground is the airspeed below which the rudder the will not be capable of producing a yawing moment. If there is an engine failure at this moment then it is must to stop the take-off run.

Right option is (d) 35 ft

Explanation: The take-off speed for the aircraft is KNOWN to be 35 feet from the ground LEVEL. The screen height is same for all TYPES of aircraft such as commercial, combat etc. This screen height is maintained same to avoid aircraft collision.

The CORRECT choice is (a) True

Explanation: The rotation speed MUST allow the aircraft to ROTATE into lift-off attitude before the lift-off speed is being achieved by the aircraft. Just before reaching lift-off speed the aircraft ROTATES into a nose-up altitude which is equal to the lift-off angle of attack.

Right answer is (a) The distance where the aircraft takes-off on the ground

To explain: The ground RUN distance is the distance where the aircraft takes-off on the ground. There are two parts of take-off distance. One is ground distance and the other one is the distance from which the aircraft leaves the ground. This SCREEN HEIGHT must be about 35 feet to avoid aircraft COLLISION.

Correct answer is (a) Maintain directional control

Explanation: The MINIMUM control SPEED airborne is used to maintain the directional control during the CLIMB of the aircraft. The minimum control speed airborne is ALWAYS greater than the minimum control speed ground.

Correct ANSWER is (d) Airspeed at which both the SAFE climb gradient and directional control is ATTAINED

The best explanation: The take-off safety SPEED is the airspeed at which both the safe climb gradient and directional control is attained. This CONDITION is attained at the time of engine failure in the airborne state.

Right choice is (d) The speed at the POINT where there is equal distances on the take-off run

Easy explanation: The decision speed is the speed at the point where there is equal distances on the take-off run. The decision speed determines the minimum SAFE LENGTH of the RUNWAY from which the aircraft can take-off.

The CORRECT ANSWER is (a) True

Easy explanation: The AIRBORNE distance is the distance between lift-off point and the point at which the screen HEIGHT is cleared. The total take-off distance is the SUM of both airborne distance and ground run distance.

The correct CHOICE is (c) The distance between lift-off POINT and the point at which the screen HEIGHT is cleared

Explanation: The airborne distance is the distance between lift-off point and the point at which the screen height is cleared. The total take-off distance is the sum of both airborne distance and ground RUN distance.

The correct answer is (d) The minimum speed at which the AIRCRAFT BECOMES airborne

For explanation I would SAY: The minimum unstick speed is the minimum speed at which the aircraft becomes airborne. This OCCURS when the over rotation pitches up the aircraft to a geometry limited angle of attack with the tail touching the GROUND.

The CORRECT CHOICE is (c) 300120 N-Km

For EXPLANATION: The ANSWER is 300120 N-Km. USE the formula ΔE=T x S. Given T=244 N and S=1230 km.

On substituting we get ΔE=244 x 1230.

On solving we get ΔE=300120 N-Km.

The CORRECT ANSWER is (d) FN-D-WsinγR-μrR=mV̇

Best EXPLANATION: The EQUATIONS derived for the take-off run are as follows:

FN-D-WsinγR-μrR=mV̇

R+L=WcosγR

Right option is (c) \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\)

Easy explanation: The formula for accelerating FORCE is given by: \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) where F is force, W is weight, μR is runway coefficient of the rolling friction, γR is runway slope, L is LIFT, CD and CL are the COEFFICIENTS of DRAG and lift.

Right answer is (a) True

Explanation: The formula for ACCELERATING force is given by: \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) where F is force, W is weight, μR is runway COEFFICIENT of the rolling friction, γR is runway slope, L is lift, CD and CL are the coefficients of drag and lift. The curly bracket in the accelerating force REPRESENTS the net propulsive thrust- weight ratio.

The correct choice is (a) True

Best explanation: The formula for accelerating force is given by: \(\frac{F}{W}=\Big\{\frac{F_N}{W}-\mu_R-sin\gamma_R\Big\}-\frac{L}{W}\Big(\frac{C_D}{C_L}-\mu_R\Big)\) where F is force, W is weight, μR is runway coefficient of the rolling friction, γR is runway slope, L is LIFT, CD and CL are the coefficients of drag and lift. The round BRACKET in the accelerating force represents the drag- lift ratio at the ground angle of ATTACK.

The correct choice is (a) True

The BEST explanation: In the calculations the horizontal airborne DISTANCE is assumed to be much greater than 35 feet. The airborne distance is given by SA=\(\frac{W}{(F_N-D)_{av}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) where

W is weight

D is drag

G is acceleration DUE to gravity

V is velocity

FN is NORMAL force.

Right choice is (c) ΔE=\(\frac{T}{S}\)

For explanation I WOULD SAY: The energy change is given by ΔE=excess thrust x distance MOVED i.e. ΔE=T x S where T is excess thrust and S is distance moved. This energy CHANCE happens when the AIRCRAFT is undergoing an airborne phase.

Correct answer is (a) True

For explanation I would say: There are TWO ATMOSPHERIC effects on the take-off distances. They are:

The take-off with REFERENCE to the indicated airspeed

The output of the power plant which is roughly proportional to relative density.

Correct choice is (d) The ground run distance is DIRECTLY proportional to AIRCRAFT drag

Easiest explanation: The correct statements are as follows:

The airborne distance is given by SA=\(\frac{W}{(F_N-D)_{AV}}\Big\{\frac{V_{2}^2-V_{LOF}^2}{2G}+35\Big\}\)

The ground run distance is given by SG=\(\frac{WV_{LOF}^2}{2g[F_N-D-\mu (W-L)]_{0.7V_{LOF}^2}}\)

The ground run distance is directly proportional to aircraft weight.

The correct option is (d) SG=\(\frac{WV_{LOF}^2}{2g[F_N-D-\mu (W-L)]_{0.7V_{LOF}^2}}\)

Explanation: The FORMULA for the ground run distance with respect to aircraft weight is given by SG=\(\frac{WV_{LOF}^2}{2g[F_N-D-\mu (W-L)]_{0.7V_{LOF}^2}}\) where W is weight, V is velocity, L is LIFT, D is drag, g is acceleration due to GRAVITY and μ is runway COEFFICIENT of ROLLING friction.

Right option is (a) True

Easy explanation: The INCREASE in aircraft weight by 10% will result in increase in aircraft ground run distance by 20%. The ground run distance is given by SG=\(\frac{WV_{LOF}^2}{2g[F_N-D-\mu (W-L)]_{0.7V_{LOF}^2}}\). From the formula it is known that the ground run distance is directly proportional to aircraft weight.

Correct answer is (a) To CHANGE the datum height

The explanation: The headwind is USED to change the datum speed of the take-off. This headwind effects the ground RUN speed. For every 10% change in headwind, there is 20% decrease in the ground run distance.

The correct answer is (a) True

Easiest explanation: The headwind only effects in the KINEMATIC energy in the case of the AIRBORNE distance. The headwind is USED to change the datum speed of the take-off. This headwind effects the GROUND RUN speed. For every 10% change in headwind, there is 20% decrease in the ground run distance.

The correct OPTION is (a) True

The best explanation: There is no effect of runway conditions on the airborne DISTANCE. The runway condition INCLUDES:

SLOPE of the runway

Friction of the runway.

The CORRECT option is (d) THRUST

The explanation: The GROUND run distance is not effected by thrust. The factors that affect ground run distance are:

TAS

EAS

Density.

Correct answer is (a) TRUE

Best EXPLANATION: The GROUND run distance is EFFECTED by the inverse of the DENSITY and directly proportional to square of the true airspeed. The ground run distance is not effected by thrust. The ground run distance is not effected by thrust. The factors that affect ground run distance are: TAS, EAS and density.

Correct answer is (a) True

The explanation: The effect of TAS on airborne distance is half that of the ideal case. The airborne distance is given by SA=\(\FRAC{W}{(F_N-D)_{AV}}\BIG\{\frac{V_{2}^2-V_{LOF}^2}{2g}+35\Big\}\) where W is weight

D is drag

G is acceleration due to gravity

V is velocity

FN is normal force.