A source emits radio signals of 100 MHz and moves towards an observer with a speed of 9Kms^–1. The apparent change in the frequency of the signals is

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3KHz

3MHz

30MHz

30KHz

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{Δ ν}{ν} = \frac{V}{C} \Rightarrow Δ ν = (\frac{V}{C}) ν \\ Δ ν = \frac{9 \times 10^{3}}{3 \times 10^{8}} \times 100 \times 10^{6} = 3000 Hz = 3 KHz \end{array}$

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