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The pitch of the whistle of an engine appears to drop by $20 \%$ of original value when it passes a stationary observer. If speed of sound in air $350 \mathrm{~m} / \mathrm{s}$, then speed of engine in $\mathrm{m} / \mathrm{s}$ is
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The correct answer is:
$87.5$
Aparent pitch heard by the observer $\mathrm{f}^{\prime}=\frac{4}{5} \mathrm{f}$
where, $f$ is the original pitch of the engine.
As the apparent frequency is lowered, thus the engine must be moving away from the stationary observer.
From Doppler effect, apparent frequency heard:
$\begin{aligned} & \mathrm{f}^{\prime}=\mathrm{f}\left[\frac{\mathrm{V}_{\text {sound }}}{\mathrm{V}_{\text {sound }}+\mathrm{V}_{\text {source }}}\right] \\ & \therefore \frac{4 \mathrm{f}}{5}=\left[\frac{350}{350+\mathrm{V}_{\text {Source }}}\right] \\ & \Rightarrow \mathrm{V}_{\text {Source }}=87.5\end{aligned}$
where, $f$ is the original pitch of the engine.
As the apparent frequency is lowered, thus the engine must be moving away from the stationary observer.
From Doppler effect, apparent frequency heard:
$\begin{aligned} & \mathrm{f}^{\prime}=\mathrm{f}\left[\frac{\mathrm{V}_{\text {sound }}}{\mathrm{V}_{\text {sound }}+\mathrm{V}_{\text {source }}}\right] \\ & \therefore \frac{4 \mathrm{f}}{5}=\left[\frac{350}{350+\mathrm{V}_{\text {Source }}}\right] \\ & \Rightarrow \mathrm{V}_{\text {Source }}=87.5\end{aligned}$
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