A) 1600 m/s
B) 1532.19 m/s
C) 160 m/s
D) zero
Correct Answer: B
Solution :
\[R=\frac{8.3}{4.2}\,cal\,g/mol/K\] \[{{C}_{V}}={{C}_{P}}-R=\left( 4.8-\frac{8.3}{4.2} \right)=2.824\] \[\gamma =\frac{{{C}_{P}}}{{{C}_{V}}}=\frac{4.8}{2.824}=1.69\] Since, \[v=\sqrt{\left( \frac{3}{\gamma } \right)}{{v}_{s}}=\sqrt{\frac{3}{1.69}}\times 1150\] \[=1532.19\,\,m/s\]
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