A) \[\frac{1}{20}s\]
B) \[\frac{1}{20}\min \]
C) \[\frac{19}{20}s\]
D) \[\frac{19}{20}\min \]
Correct Answer: C
Solution :
From Dopplers effect in sound the apparent change in frequency of the source due to a relative motion between source and observer is \[n=n\left( \frac{v-{{v}_{o}}}{v-{{v}_{s}}} \right)\] Given, \[{{v}_{o}}=0,\,\,{{v}_{s}}=\frac{v}{20}\] \[n=1\,\,Hz\](as blast is blown at an interval of\[1s\]) \[\therefore \] \[n=\frac{v}{v-\frac{v}{20}}\times 1=\frac{20}{19}Hz\] Observed time interval between two successive blasts\[=\frac{19}{20}s\].
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