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DGCA Technical General · 55 questions

Mach: DGCA Technical General Questions and Answers

55 practice questions on Mach with answers and explanations, plus the key concepts and formulas, from the TrueHeading question bank.

About this topic

Use this page to revise Mach for the DGCA Technical General paper. Below are the key ideas first, then 12 practice questions with answers and explanations, picked from the 55 questions in the TrueHeading bank for this topic.

In short: The speed of sound depends only on temperature.

Open the full set in the student zone to answer every question, track your accuracy and take timed mock tests.

Key concepts: Mach

4 ideas to know cold.

Local speed of sound and Mach number?

The speed of sound depends only on temperature.

LSS=38.95T(K) kt\text{LSS}=38.95\sqrt{T(\text{K})}\ \text{kt}
M=TASLSSM=\frac{TAS}{\text{LSS}}
  • ISA sea level: 661 kt
  • At or above 36 090 ft: 573 kt (−56.5 °C)
  • Mach angle: sin μ = 1 / M
μMach cone: sin μ = 1 / Mabove Mach 1 the pressure waves pile up into a shock wave

Critical Mach and compressibility effects?

MCRIT is the free-stream Mach number at which local airflow first reaches Mach 1.

  • Shock wave, wave drag, shock stall, Mach tuck
  • Sweepback and thin wings raise MCRIT
  • Mach trim compensates for tuck

What happens to IAS and Mach in climb and descent?

In a constant-IAS climb, Mach increases. In a constant-Mach climb, IAS decreases.

  • At the crossover altitude the speed reference changes from IAS to Mach (about FL 280–310)
  • VMO is an IAS limit, MMO a Mach limit

Define Mach number and the speed of sound.

M = TAS / LSS, where LSS = 38.95 √T kt (T in kelvin).

M=TASa,a=38.95TM=\dfrac{TAS}{a},\quad a=38.95\sqrt{T}
  • LSS at sea level ISA ≈ 661 kt
  • LSS at ISA tropopause ≈ 573 kt
  • Mcrit: first local M=1
  • Mmo: operating limit

Practice questions: Mach

Tap “Show answer” after you have tried each one.

Q1At an outside air temperature of −30 °C an aircraft flies at Mach 0.74. Using LSS = 38.95 √T (kt, T in K), the TAS is approximately (kt):

  1. 430
  2. 465
  3. 480
  4. 450
Show answer

Answer: D. 450

LSS = 38.95 × √243.1 ≈ 607 kt; TAS = Mach × LSS ≈ 449 kt.

Q2At an outside air temperature of 15 °C an aircraft flies at Mach 0.74. Using LSS = 38.95 √T (kt, T in K), the TAS is approximately (kt):

  1. 505
  2. 490
  3. 470
  4. 520
Show answer

Answer: B. 490

LSS = 38.95 × √288.1 ≈ 661 kt; TAS = Mach × LSS ≈ 489 kt.

Q3At an outside air temperature of −65 °C an aircraft flies at Mach 0.74. Using LSS = 38.95 √T (kt, T in K), the TAS is approximately (kt):

  1. 415
  2. 430
  3. 445
  4. 395
Show answer

Answer: A. 415

LSS = 38.95 × √208.1 ≈ 562 kt; TAS = Mach × LSS ≈ 416 kt.

Q4At an outside air temperature of −56 °C an aircraft flies at Mach 0.84. Using LSS = 38.95 √T (kt, T in K), the TAS is approximately (kt):

  1. 495
  2. 510
  3. 460
  4. 480
Show answer

Answer: D. 480

LSS = 38.95 × √217.1 ≈ 574 kt; TAS = Mach × LSS ≈ 482 kt.

Q5At an outside air temperature of −56 °C an aircraft flies at Mach 0.78. Using LSS = 38.95 √T (kt, T in K), the TAS is approximately (kt):

  1. 450
  2. 480
  3. 465
  4. 430
Show answer

Answer: A. 450

LSS = 38.95 × √217.1 ≈ 574 kt; TAS = Mach × LSS ≈ 448 kt.

Q6TAS is 380 kt and the outside air temperature is 0 °C. Using LSS = 38.95 √T, the Mach number is approximately:

  1. 0.52
  2. 0.59
  3. 0.65
  4. 0.71
Show answer

Answer: B. 0.59

LSS ≈ 644 kt; Mach = 380/644 ≈ 0.590.

Q7TAS is 340 kt and the outside air temperature is −50 °C. Using LSS = 38.95 √T, the Mach number is approximately:

  1. 0.70
  2. 0.51
  3. 0.64
  4. 0.58
Show answer

Answer: D. 0.58

LSS ≈ 582 kt; Mach = 340/582 ≈ 0.584.

Q8TAS is 460 kt and the outside air temperature is −56 °C. Using LSS = 38.95 √T, the Mach number is approximately:

  1. 0.86
  2. 0.73
  3. 0.92
  4. 0.80
Show answer

Answer: D. 0.80

LSS ≈ 574 kt; Mach = 460/574 ≈ 0.801.

Q9TAS is 380 kt and the outside air temperature is −40 °C. Using LSS = 38.95 √T, the Mach number is approximately:

  1. 0.70
  2. 0.57
  3. 0.64
  4. 0.76
Show answer

Answer: C. 0.64

LSS ≈ 595 kt; Mach = 380/595 ≈ 0.639.

Q10The Mach number is defined as:

  1. IAS divided by the speed of sound
  2. TAS divided by the speed of sound at sea level
  3. TAS divided by the local speed of sound
  4. CAS divided by TAS
Show answer

Answer: C. TAS divided by the local speed of sound

M = TAS / LSS.

Q11In a climb at a constant IAS, the Mach number:

  1. stays constant
  2. is zero
  3. increases
  4. decreases
Show answer

Answer: C. increases

TAS increases and the LSS decreases.

Q12The Mach meter is basically:

  1. a pitot probe alone
  2. a barometer
  3. an airspeed indicator combining pitot-static pressures with altitude data
  4. a gyroscope
Show answer

Answer: C. an airspeed indicator combining pitot-static pressures with altitude data

It senses the ratio of dynamic to static pressure.

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