A) \[951m/sec\]
B) \[159m/sec\]
C) \[591m/sec\]
D) \[519m/sec\]
Correct Answer: C
Solution :
[c]| we have \[\frac{{{v}_{m}}}{{}^{v}H}=\sqrt{\frac{{{\rho }_{H}}}{{{\rho }_{m}}}}\] | ? (1) |
| Here, \[{{\rho }_{m}}=\frac{{{m}_{H}}+{{m}_{N}}}{{{V}_{H}}+{{V}_{N}}}\] | |
| \[=\frac{{{\rho }_{H}}{{V}_{H}}+{{\rho }_{N}}{{V}_{N}}}{{{V}_{H}}+{{V}_{N}}}\] | |
| \[=\frac{{{\rho }_{H}}{{V}_{H}}\left( 1+\frac{{{\rho }_{N}}}{{{\rho }_{H}}}\times \frac{{{V}_{N}}}{{{V}_{H}}} \right)}{{{V}_{H}}\left( 1+\frac{{{V}_{N}}}{+{{V}_{H}}} \right)}\] | |
| \[\therefore \frac{{{\rho }_{m}}}{{{\rho }_{H}}}=\frac{1+\frac{28}{2}\times \frac{1}{2}}{1+\frac{1}{2}}=\frac{16}{3}\] | |
| Now from equation (i), we have \[{{[{{v}_{m}}]}_{0}}={{v}_{H}}=\sqrt{\frac{3}{16}}\] | |
| \[\frac{1300\sqrt{3}}{4}=325\sqrt{3}\] | |
| \[{{[{{v}_{m}}]}_{27}}={{({{v}_{m}})}_{0}}\left( 1+\frac{t}{546} \right)=591m/s\] | |
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