A) \[\frac{1}{20}S\]
B) \[1/20\,min\]
C) \[19/20\text{ }S\]
D) \[19/20\,min\]
Correct Answer: C
Solution :
According to Doppler effect in sound the app change in frequency of the source due to a relative motion between source and observer is: \[n'=n\left( \frac{V-{{V}_{0}}}{V-{{V}_{S}}} \right)\] Where V is velocity of sound, \[{{V}_{0}}\] the velocity of observer and \[{{V}_{S}}\] the velocity of source. Here, \[{{V}_{0}}=0\] (observer stationary) \[{{V}_{S}}=\frac{V}{20}\] \[n=1Hz.\] (as blast in blown at an interval of Is) \[n'=\frac{V}{V-\frac{V}{20}}\times 1=\frac{20}{19}Hz.\] Observed time interval between two successive blasts \[=\frac{19}{20}S.\]You need to login to perform this action.
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