Answer :

Correct answer is : 374.77

V = 300 m/s, T = 330 K, γ = 1.4, R = 287 J/kg-K.

$M=\frac{V}{\sqrt{\gamma RT}}=\frac{300}{\sqrt{1.4\times 287\times 330}}$ = 0.823

We know that,

$\frac{T_0}{T}=1+\left(\frac{\gamma -1}{2}\right)M^2$

$T_0=T\{1+\left(\frac{1.4-1}{2}\right){0.823}^2\}$

T_o = 330 (1 + 0.1355) = 374.71 K

∴ Stagnation temperature T_o = 374.71 K