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Right choice is (d) 0

To explain: In gliding FLIGHT the VALUE of dimensionless power (λ) is zero (0) and the SHALLOW glide ANGLE is given by flying at the MINIMUM drag speed. As the power increase the airspeed for maximum climb gradient decreases.

Right choice is (c) SC=\(\frac{dH/dt}{Q_f}\)c

Easiest EXPLANATION: The FORMULA for specific climb is GIVEN by SC=\(\frac{dH/dt}{Q_f}\) where SC is specific thrust, \(\frac{dH}{dt}\) is RATE change of height with respect to time and Qf is fuel flow. The specific climb is CALCULATED in the units of ft/kg.

Right choice is (c) J=\(\frac{V}{nD}\)

The explanation: The advanced ratio is GIVEN by J=\(\frac{V}{nD}\) where J is advance ratio, V is velocity and D is PROPELLER DIAMETER and n is rotational speed. Propeller efficiency is depended on advanced ratio.

Correct CHOICE is (a) the aircraft flying at airspeed less than minimum drag speed

The EXPLANATION: The flight PATH is unstable during the aircraft flying at airspeed less than minimum drag speed. In this CASE the minimum rate of descent occurs at the minimum power speed and minimum gradient occurs at the minimum drag speed.

Correct answer is (a) air MASS

For explanation: The flight path is calculated relative to air mass. This resulted in the development of the THEORY by experimenting in the still air condition. In actual performance of the aircraft the relative velocity is not EFFECTED by the REFERENCE axis as they are velocity axes.

Correct ANSWER is (a) True

Best explanation: WAT anticipated changes result in better CLIMB PERFORMANCE. The thrust required during the take-off is a FUNCTION of weight, altitude and temperature i.e. WAT. Thrust is directly proportional to airspeed.

Correct choice is (a) Climb rate increases at minimum POWER drag

To elaborate: Climb rate is maximum at minimum power drag. The rate of climb is given by [ηP-DV]\(\frac{1}{W}\)=\(\frac{DH}{DT}\) where η is propeller efficiency, P is power, V is velocity, D is drag and W is weight and dH/dt is vertical velocity.

The correct choice is (a) True

To explain I WOULD SAY: The relation between power and airspeed for climb gradient is that the power increases with decrease in airspeed for MAXIMUM climb gradient. The best climb gradient is ATTAINED by flying at airspeeds LESS than the minimum power speeds.

The CORRECT choice is (c) After take-off path

To explain: The most crucial and CRITICAL path in the flight path is the after take-off path. This is OBTAINED by extending landing gears and flaps set to optimize take-off speed and runway distance requirement.

The correct answer is (b) OPTIMUM climb speed

Easiest explanation: During the calculated fuel flow flight trials the optimum climb speed is calculated and then the SPECIFIC climb function is known which HELPS in the determination of airspeed for a MINIMUM fuel climb.

Right answer is (b) nose uppitch

To explain: The climb GRADIENT is decreased by decrease in nose down pitch. This can be ALONE done by the ELEVATOR control. This is the SITUATION of the airspeed less than minimum drag speed. If the aircraft is flying with airspeed greater than minimum drag speed then the FLIGHT path gradient of descent can be increased by increasing airspeed.

Right answer is (c) \(\Big[\frac{\eta P}{V}-D\Big]\frac{1}{W}\)=sinγ2

Best explanation: The gradient of climb is GIVEN by \(\Big[\frac{\eta P}{V}-D\Big]\frac{1}{W}\)=sinγ2 where η is propeller EFFICIENCY, P is power, V is velocity, D is drag and γ is the angle at which the force is acting on the aircraft. The value of gradient of climb is more when the propulsive force is maximum.

Right CHOICE is (d) airspeed below minimum drag speed

To ELABORATE: The best gradient of climb is attained at airspeed below minimum drag speed. The best rate of climb is attained at airspeed above minimum drag speed. The ACTUAL AIRSPEEDS will be a function of excess propulsive thrust.

The correct answer is (a) u=0.76

The best I can EXPLAIN: The maximum RATE of climb OCCURS at the airspeed u=0.76 m/sec. The best gradient of climb is attained at airspeed below minimum drag speed. The best rate of climb is attained at airspeed above minimum drag speed. The actual airspeeds will be a function of excess propulsive thrust.

Right answer is (a) True

Best explanation: Drag raise is avoided by maintaining constant mach NUMBER. The best economic flight is ACHIEVED by minimum drag speed and TAKE minimum TIME to climb. The best rate of climb is achieved by LESS drag raise.

Right choice is (a) True

For explanation I would say: The FORMULA for SPECIFIC climb is given by SC=\(\frac{dH/dt}{Q_f}\) where SC is specific thrust, \(\frac{dH}{dt}\) is RATE change of height with RESPECT to time and Qf is fuel flow. The specific climb is CALCULATED in the units of ft/kg.

The correct ANSWER is (d) 300 ft/min

The best I can explain: The rate of the change in the CABIN pressure should not exceed 300 ft/min at sea LEVEL. This STATES that the cabin is pressurized to 8000ft pressure height, the descent to the sea level should not exceed 24 minutes before the take-off.

Right choice is (a) decreases the wind speed with decrease in height

To EXPLAIN: The RELATIVE velocity of the wind produces BOUNDARY layer close to the ground which decreases the wind speed with decrease in height. This will further produce effects on the aircraft.

Correct choice is (B) False

Easiest explanation: It is assumed zero velocity datum at the CENTER of gravity (CG) of the AIRCRAFT. Centre of gravity (CG) of an aircraft is the point at which the whole WEIGHT of the aircraft is being balanced. This CG does not change with respect to SPEED of the aircraft.

Right CHOICE is (B) 15000m

Easy explanation: The safe height of an aircraft for take-off above the airfield is 15000m. The USE of safe height is to avoid ground BASED obstacles and make a clear flight PATH to avoid aircraft collision and loss of life and property.

Right choice is (c) Power increases with decrease in airspeed

For explanation: The RELATION between power and airspeed for climb gradient is that the power increases with decrease in airspeed for MAXIMUM climb gradient. The best climb gradient is ATTAINED by flying at AIRSPEEDS less than the minimum power speeds.

The correct choice is (B) False

Easiest explanation: The BEST gradient of climb is attained at the time of minimum drag SPEED of the AIRCRAFT. The best gradient of climb is attained generally at the airspeed ‘U’ of 1. The gradient of climb is given by Emaxsinγ2=τ-\(\frac{1}{2}\)[u^2+u^-2].

Right ANSWER is (B) False

Best explanation: The airspeed chosen for after take-off climb should be same as for that of maximum gradient.The safeguard path has a gradient of 2%. The safeguard DISTANCE EXTENDS from the take-off distance available on the runway to a distance of 15000m.

The CORRECT option is (B) False

The explanation is: The best rate of CLIMB occurs when the excess thrust power (FNV) is maximum than the drag power (DV). As the ideal power increases linearly with true airspeed the best gradient of climb is predicted to be at an airspeed greater than minimum drag speed.

The CORRECT option is (a) Safe margining over the stall airspeed

To explain I would say: The RESTRICTIONS on the airspeed scheduled for climb gradient are safe margining over the stall airspeed and the ability to maintain lateral-directional control over the flight during the LOSS of PROPULSIVE thrust engine.

Right OPTION is (c) WEIGHT

Easiest explanation: The thrust required during the take-off is a function of weight, altitude and temperature i.e. WAT. Thrust is directly proportional to airspeed. This means the thrust increases with increase in airspeed and decrease with decrease in decrease in airspeed. The airspeed best for RATE of climb to close to that of their MINIMUM drag speed.

Correct choice is (a) True

Explanation: The minimum fuel CLIMB occurs at the AIRSPEED for best rate of climb. During the calculated fuel flow flight trials the optimum climb speed is calculated and then the specific climb FUNCTION is known which HELPS in the determination of airspeed for a minimum fuel climb.

Right choice is (b) considering horizontal component of the wind

The explanation: The AIRCRAFT is taken to the level ground OPERATING for considering horizontal component of the wind. Wind is the relative velocity between the air mass and the ground. Usually wind is assumed to have only a horizontal component but there exists VERTICAL component which MAY AFFECT the aircraft at considerable height.

The correct choice is (a) True

The best I can explain: The flight path can be CONTROLLED by the use of elevator control only. If the aircraft is FLYING with airspeed greater than minimum drag speed then the flight path gradient of descent can be INCREASED by INCREASING airspeed.

Correct answer is (C) Elevator and thrust settings

To explain I would SAY: The precise CONTROL of the flight PATH gradient is achieved by the changes in the thrust of the flight along with the elevator control inputs given to the system during the gradient of descent of the AIRCRAFT.

The correct answer is (a) True

Easiest explanation: The simple rule THUMB helps in maintain good relation between RATE of descent and airspeed. This strategy can be IMPROVED by implementing the flight MANAGEMENT system in the descent flight PATH gradient.

Right CHOICE is (d) Visualization of the aircraft with RESPECT to any one frame such as aircraft or GROUND

Easiest explanation: Visualization of the aircraft with respect to any one frame such as aircraft or ground is KNOWN as the PERCEIVED performance. This effects the ability of the aircraft to clear ground-based obstructions.

Correct choice is (a) 1

To elaborate: The maximum GRADIENT of climb occurs at AIRSPEED (U) =1. There occurs negative values for sinγ2of τ and u when the flight is descending. The climb gradient equation is GIVEN by the equation τ-\(\frac{1}{2}\)[u^2+u^-2]=Emaxsinγ2.

The correct answer is (b) [ηP-DV]\(\frac{1}{W}\)=\(\frac{dH}{dt}\)

The best I can explain: The RATE of CLIMB is given by [ηP-DV]\(\frac{1}{W}\)=\(\frac{dH}{dt}\) where η is propeller EFFICIENCY, P is power, V is velocity, D is drag and W is WEIGHT and dH/dt is vertical velocity. The rate of climb is maximum when power speed is minimum.

The correct OPTION is (a) ν=\(\frac{dH/dt}{V_{md}}\)

The BEST I can explain: The correct formula for dimensionless rate of climb is given by the formula V=\(\frac{dH/dt}{V_{md}}\) where v the dimensionless rate of climb is and dH/dt is VERTICAL VELOCITY and Vmd is velocity. The vertical velocity dH/dt is given in feet/min.

The correct answer is (a) True

For explanation I would say: The AIRSPEED best for RATE of climb to CLOSE to that of their minimum drag speed. The operating height is ACHIEVED by MAXIMUM rate climb. The safe height of an aircraft for take-off above the airfield is 15000m. The use of safe height is to avoid ground based obstacles and make a clear flight path to avoid aircraft collision and loss of life and property.

Right ANSWER is (a) True

The explanation is: The flight MACH number is inversely proportional to relative pressure. This STATES that the flight mach number increases with DECREASE in relative pressure and decreases WIT increase in relative pressure.

The correct option is (b) False

The explanation: The performance relative to ground is effected by the wind. The relative velocity of the wind produces boundary layer CLOSE to the ground which DECREASES the wind SPEED with decrease in height. This will further produce effects on the aircraft.

The correct choice is (b) False

To elaborate: The rate of climb or descent relative to air MASS is unaffected by the wind but the horizontal COMPONENT of true airspeed is EFFECTED by the tailwind. The effect is increased by tailwind and decreased by the headwind.

The correct answer is (a) INCREASED by tailwind

The BEST explanation: The rate of CLIMB or descent RELATIVE to air mass is unaffected by the wind but the horizontal component of true airspeed is EFFECTED by the tailwind. The effect is increased by tailwind and decreased by the headwind.

The CORRECT CHOICE is (d) 400kts/min

Easy explanation: If ETS is 400kts, then the ASSOCIATED acceleration with it is 400kts/min. This is during the climb phase of the aircraft. Thus it concludes that the climb is the combination of the potential and KINETIC ENERGY.

The CORRECT option is (d) Specific TOTAL energy

For EXPLANATION: Specific total energy is the potential and kinetic energy per unit mass. This is few times referred as the energy height as it represents the height of the aircraft would ATTAIN if all the kinetic energy is converted to potential energy.

Correct option is (d) Es=\(\Big\{H+\frac{V^2}{2G}\Big\}\)

For explanation I would say: The correct formula for specific total energy is given by the formula Es=\(\Big\{H+\frac{V^2}{2g}\Big\}\) where ‘Es’ is specific total energy, ‘H’ is height, ‘V’ is VELOCITY and ‘g’ is acceleration DUE to gravity. Specific total energy is the potential and kinetic energy per unit mass. This is few times REFERRED as the energy height as it represents the height of the AIRCRAFT would attain if all the kinetic energy is converted to potential energy.

Correct ANSWER is (b) Specific EXCESS Power

Easy explanation: SEP stands for specific excess power. It is equal to the rate of change of specific TOTAL energy (STE) and is used to increase the total energy of the aircraft which is in a combination with CLIMB and acceleration.