A source and a detector move away from each other in absence of wind with a speed of $20\, {m} / {s}$ with respect to the ground. If the detector detects a frequency of $1800\, {Hz}$ of the sound coming from the source, then the original frequency of source considering speed of sound in air $340\, {m} / {s}$ will be ${Hz}$
JEE MAIN 2021, Diffcult
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${V}_{{S}}=20 {m} / {s} \quad {V}_{{O}}=20 {m} / {s}$
${f}^{\prime}={f}\left(\frac{{V}-{V}_{0}}{{V}+{V}_{{s}}}\right)$
$1800={f}\left(\frac{340-20}{340+20}\right)$
${f}=2025 {Hz}$
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