A) 6 m/sec
B) 3 m/sec
C) 1.5 m/sec
D) 12 m/sec
Correct Answer: B
Solution :
When the source approaches the observer Apparent frequency \[n'=\frac{v}{v-{{v}_{s}}}.n=n\left[ \frac{1}{1-\frac{{{v}_{s}}}{v}} \right]\] =\[n{{\left[ 1-\frac{{{v}_{s}}}{v} \right]}^{-1}}=n\left[ 1+\frac{{{v}_{s}}}{v} \right]\] (Neglecting higher powers because vS << v) When the source recedes the observed apparent frequency \[{n}''=n\left[ 1-\frac{{{v}_{s}}}{v} \right]\] Given \[n'-n''=\frac{2}{100}n,\,\,v=300\]\[m/\sec \] \[\therefore \] \[\frac{2}{100}n=n\left[ 1+\frac{{{v}_{s}}}{v} \right]-n\left[ 1-\frac{{{v}_{s}}}{v} \right]=n\left[ 2\frac{{{v}_{s}}}{v} \right]\] \[\Rightarrow \] \[\frac{2}{100}=2\frac{{{v}_{s}}}{v}\Rightarrow {{v}_{s}}=\frac{v}{100}=\frac{300}{100}=3\]\[m/\sec \].
You need to login to perform this action.
You will be redirected in
3 sec
