A) 20 m/s
B) 10 m/s
C) 30 m/s
D) 33 m/s
Correct Answer: C
Solution :
From Dopplers/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 Eqs. (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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