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On Day 1, an aircraft flies with a speed of V1 m/s at an altitude where the temperature is T1 K. On Day 2, the same aircraft flies with a speed of √1.2 * V1 m/s at an altitude where the temperature is 1.2 * T1 K. How does the Mach number M2 on Day 2 compare with the Mach number M1 on Day 1? Assume ideal gas behavior for air. Also assume the ratio of specific heats and molecular weight of air to be the same on both days.
A
M2 = 0.6 * M1
B
M2 = M1
C
M2 = (1/√1.2) * M1
D
M2 = √1.2 * M1

Correct : b

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