A) \[\frac{1}{20}\] sec
B) \[\frac{1}{20}\] min
C) \[\frac{19}{20}\] sec
D) \[\frac{19}{20}\] min
Correct Answer: C
Solution :
The blast is blows at an interval of 1 sec, so frequency\[=1\text{ }Hz\]. \[\therefore \] Frequency heard by the observer \[{{n}^{1}}=\frac{\upsilon }{\upsilon -{{\upsilon }_{s}}}.n\] \[=\frac{\upsilon }{\upsilon -\frac{\upsilon }{20}}\times 1\] \[=\frac{\upsilon }{19\frac{\upsilon }{20}}=\frac{20}{19}Hz\] Therefore, observed time interval between two successive blasts \[=\frac{1}{20/19}=\frac{19}{20}\sec \] \[\therefore \] \[C\propto R\]You need to login to perform this action.
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