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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