A) \[1092{}^\circ C\]
B) 819 K
C) \[819{}^\circ C\]
D) \[546{}^\circ C\]
Correct Answer: C
Solution :
Speed \[v\propto \sqrt{T}\] \[\therefore \] \[\frac{2v}{v}=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\] \[\Rightarrow \] \[2=\sqrt{\frac{{{T}_{2}}}{{{T}_{1}}}}\] \[\Rightarrow \] \[4=\frac{{{T}_{2}}}{{{T}_{1}}}\] \[\Rightarrow \] \[{{T}_{2}}=4{{T}_{1}}=4\times 273\] \[{{T}_{2}}=1092\,K\] \[\Rightarrow \] \[{{T}_{2}}=1092-273=819{{\,}^{o}}C\]
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