A) \[20\,\,m/s\]
B) \[10\,\,m/s\]
C) \[30\,\,m/s\]
D) \[33\,\,m/s\]
Correct Answer: C
Solution :
From Doppler's effect, the perceived frequency when source approaches observer is \[n'=n\left( \frac{v}{v-{{v}_{s}}} \right)\] ? (i) When source recedes the observer \[n'\,\,'=n\left( \frac{v}{v+{{v}_{s}}} \right)\] ... (ii) From Eq. (i) and (ii), we get \[\frac{n'}{n'\,\,'}=\frac{v+{{v}_{s}}}{v-{{v}_{s}}}\] \[\frac{6}{5}=\frac{330+{{v}_{s}}}{330-{{v}_{s}}}\] \[\Rightarrow \] \[1980-6{{v}_{s}}=1650+5{{v}_{s}}\] \[\Rightarrow \] \[11{{v}_{s}}=1980-1650=330\] \[{{v}_{s}}=\frac{330}{11}=30\,\,m/s\]
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