A) 360
B) 330
C) 320
D) 340
Correct Answer: A
Solution :
When a listener moves towards a stationary source apparent frequency \[{n}'=\left( \frac{v+{{v}_{0}}}{v} \right)\,\]\[n=200\] ?..(i) When listener moves away from the same source \[{n}''=\frac{(v-{{v}_{O}})}{v}n=160\] ?..(ii) From (i) and (ii) \[\frac{v+{{v}_{O}}}{v-{{v}_{O}}}=\frac{200}{160}\]Þ \[\frac{v+{{v}_{O}}}{v-{{v}_{O}}}=\frac{5}{4}\]Þ \[v=360m/\sec \]You need to login to perform this action.
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