A) 2.0 m/s
B) 2.5 m/s
C) 3.0 m/s
D) 3.5 m/s
Correct Answer: B
Solution :
Apparent frequency due to source A \[n=\frac{v-{{v}_{0}}}{v}\] \[=\frac{v-4}{v}\times n\] Apparent frequency due to source B \[n=\frac{v+u}{v}\] \[=\frac{v-u}{v}\times n\] \[\therefore \] \[n-n=\frac{2u}{v}\times n={{I}_{0}}\] \[u=\frac{10u}{2n}\] \[=\frac{10\times 340}{2\times 680}\] \[=2.5\,m/s\]You need to login to perform this action.
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