A) 33 m/sec
B) 47 m/sec
C) 56 m/sec
D) 63 m/sec
Correct Answer: A
Solution :
\[n'=n\left( \frac{v-{{v}_{0}}}{v-{{v}_{s}}} \right)\] Given: \[\frac{n'}{n}=\frac{90}{100}\] \[v=330\,\,m/s\] and\[{{v}_{s}}=0\] \[\therefore \] \[\frac{90}{100}=\frac{330-{{v}_{0}}}{330-0}\] \[\Rightarrow \] \[330-{{v}_{0}}=\frac{99}{100}\times 330=297\] \[\Rightarrow \] \[{{v}_{0}}=330-297=33\,\,m/\sec \]
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