MHT CET · Physics · Waves and Sound
When an observer moves towards a stationary source with velocity ' \(\mathrm{V}_1\) ', the appearent frequency of emitted note is ' \(\mathrm{F}_1\) '. When observer moves away from stationary source with velocity ' \(\mathrm{V}_1\) ' the appearent frequency is ' \(F_2\) '. If ' \(v\) ' is velocity of sound in air and \(\frac{F_1}{F_2}=2\), then \(\frac{V}{V_1}\) is equal to
- A \(6\)
- B \(5\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(C) \(3\)
Step-by-step Solution
Detailed explanation
\(F_1 = F_0 \left( \frac{v + V_1}{v} \right)\) \(F_2 = F_0 \left( \frac{v - V_1}{v} \right)\)
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