A) zero
B) 5%
C) 10%
D) 0.1%
Correct Answer: C
Solution :
When the observer is moving towards a stationary source Apparent frequency \[v'=v\left[ \frac{v+{{v}_{0}}}{v} \right]\] Given, \[{{v}_{0}}=\frac{v}{10}\] \[v'=v\left[ \frac{v+\frac{v}{10}}{v} \right]\] \[v'=\frac{11}{10}v\] \[\therefore \] Change in frequency \[=\frac{11}{10}v-v\] \[=\frac{1}{10}v\] \[\therefore \] % change =10%
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