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 the days.
Options:
(A) $M_2 = 0.6 M_1$
(B) $M_2 = M_1$
(C) $M_2 = \frac{1}{\sqrt{1.2}} M_1$
(D) $M_2 = \sqrt{1.2} M_1$
Official Answer Key:
B
Step-by-Step Solution & Explanation
### **Detailed Step-by-Step Solution (GATE AE 2024 Q.22)**
**1. Flight Parameters on Day 1 and Day 2**:
- **Day 1**: Flight speed $V_1$, ambient temperature $T_1$.
- **Day 2**: Flight speed $V_2 = \sqrt{1.2}\,V_1$, ambient temperature $T_2 = 1.2\,T_1$.
- Ratio of specific heats $\gamma$ and molecular weight (and thus gas constant $R$) are identical on both days.
**2. Speed of Sound and Mach Number Derivation**:
The speed of sound in an ideal gas is given by:
$$a = \sqrt{\gamma R T}$$
The flight Mach number is defined as $M = \frac{V}{a} = \frac{V}{\sqrt{\gamma R T}}$.
Evaluating for Day 2:
$$M_2 = \frac{V_2}{\sqrt{\gamma R T_2}} = \frac{\sqrt{1.2}\,V_1}{\sqrt{\gamma R (1.2\,T_1)}} = \frac{\sqrt{1.2}\,V_1}{\sqrt{1.2}\sqrt{\gamma R T_1}} = \frac{V_1}{\sqrt{\gamma R T_1}} = M_1$$
**3. Conclusion**:
$$M_2 = M_1$$
**4. Official Master Key**:
- **Official Answer Key**: **B** ($M_2 = M_1$)
- **Exam Takeaway**: Since both flight speed and speed of sound scale with $\sqrt{1.2}$, their ratio (the Mach number) remains invariant.