250+ TOP MCQs on Flight Mechanics – Steady Climbing and Descending Flight and Answers

Aircraft Design written test Questions & Answers on “Flight Mechanics – Steady Climbing and Descending Flight”.

1. What do you mean by rate of climb?
a) Vertical velocity of an aircraft
b) Lift of an aircraft
c) Thrust required
d) Drag polar
Answer: a
Clarification: Rate of climb is nothing but the vertical velocity of aircraft. It is the climbing rate of an aircraft. Lift is an aerodynamic force which holds the aircraft in the air. Weight is the force acting due to gravity. Thrust is propulsive force.

2. Ratio of vertical distance travelled to the horizontal distance travelled is known as __________
a) climb gradient
b) lift curve slope
c) power required
d) thrust loss
Answer: a
Clarification: Ratio of vertical distance travelled to the horizontal distance travelled is known as Climb gradient. Lift curve slope is change lift coefficient to the change in angle of attack. Power required is product of Thrust required and velocity of the aircraft.

3. Which of the following is correct in terms of fate of climb (R/C)?
a) R/C = excess power/weight
b) R/C = lift/drag
c) R/C = excess lift/drag
d) R/C = excess power/Thrust required
Answer: a
Clarification: Rate of climb is nothing but the vertical velocity of the aircraft. Rate of climb can be defined as the ratio of excess power of the aircraft to the aircraft weight. Hence, Correct relation between rate of climb and power is as follows: R/C = excess power/weight. Where, R/C = rate of climb.

4. Consider the vertical velocity of the aircraft is 10m/s and horizontal velocity is 12 m/s. Determine the value of climb gradient.
a) 0.833
b) 1.89
c) 8
d) 2.483
Answer: a
Clarification: Climb gradient = vertical velocity/ horizontal velocity = 10/12 = 0.833.

5. Find lift to weight ratio if climb angle is 45°.
a) 0.707
b) 1
c) 1.34
d) 0.992
Answer: a
Clarification: Lift to weight ratio = cos(climb angle) = cos(45°) = 0.707.

6. Determine the value of climb angle if, excess thrust is 40 unit and weight of the aircraft is 60 unit. Consider steady climb.
a) 41.8
b) 50
c) 78
d) 12
Answer: a
Clarification: Climb angle = arcsine (excess Thrust/weight) = arcsine (40/60) = 41.8°.

7. Find the approximate value of climb angle if Thrust is 1500N, drag is 1000N and weight of the aircraft is 2500N.
a) 11.53 degree
b) 30 degree
c) 40 degree
d) 1 degree
Answer: a
Clarification: Given, steady climb, Thrust=1500N, drag D=1000N, weight W=2500N.
Climb angle = arcsine [{T-D}/W]
= arcsine [{1500-1000}/2500] = 11.53 degree.

8. Determine the value of fuel burned during steady climb if SFC C is 0.000029 per second, thrust is 150KN and time required is 1500s.
a) 6.525 KN
b) 100N
c) 345
d) 87.87 N
Answer: a
Clarification: Fuel burned = SFC*Thrust*Time
= 0.000029*150*1500*1000 = 6.525KN.

9. Find the approximate value of velocity of best climb. Consider Thrust loading as 1.08, wing loading as 377 unit, CD0 as 0.025, K as 0.0075 and sea level density as 1.225 unit.
a) 98.46
b) 37.45
c) 12.1
d) 189.982