A) \[\frac{1}{3}\]
B) \[\frac{3}{10}\]
C) \[\frac{2}{5}\]
D) \[\frac{2}{3}\]
Correct Answer: B
Solution :
Given, \[{{v}_{s}}=30\,m{{s}^{-1}}\] \[{{v}_{o}}=60\,m{{s}^{-1}}\] Apparent frequency \[v'=v\left[ \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right]\] \[v'=v\left[ \frac{330+60}{330-30} \right]\] \[v'=v\left[ \frac{390}{300} \right]\] \[\Rightarrow \] s\[v'=\frac{39}{30}v\] \[\therefore \] Fractional change in frequency \[\frac{\frac{39}{30}v-v}{v}=\frac{9}{30}=\frac{3}{10}\]You need to login to perform this action.
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