Question

Day 1: speed \(V_1\) at temperature \(T_1\) K. Day 2: speed \(\sqrt{1.2}\,V_1\) at temperature \(1.2\,T_1\) K. For air treated as an ideal gas with the same \(\gamma\) and \(R\) both days, compare Mach numbers \(M_2\) (Day 2) and \(M_1\) (Day 1).

• A. \(M_2=0.6\,M_1\)
• B. \(M_2=M_1\)
• C. \(M_2=\dfrac{1}{\sqrt{1.2}}\,M_1\)
• D. \(M_2=\sqrt{1.2}\,M_1\)

Hint:

If both \(V\) and \(a\) scale by the same factor, Mach number stays unchanged. Here both scale by \(\sqrt{T}\).

Answer 1

The Correct Option is B

Step 1: Mach number definition.
\(M=\dfrac{V}{a}\), with speed of sound \(a=\sqrt{\gamma R T}\). 

Step 2: Evaluate each day.
Day 1: \(M_1=\dfrac{V_1}{\sqrt{\gamma R T_1}}\).
Day 2: \(a_2=\sqrt{\gamma R (1.2T_1)}=\sqrt{1.2}\,a_1\), and \(V_2=\sqrt{1.2}\,V_1\). Therefore \[ M_2=\frac{V_2}{a_2} =\frac{\sqrt{1.2}\,V_1}{\sqrt{1.2}\,a_1} =\frac{V_1}{a_1} =M_1. \] \[\boxed{M_2=M_1}\]