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A speaker emits a sound wave of frequency $f_{0}$. When it moves towards a stationary observer with speed $u$, the observer measures a frequency $f_{1}$. If the speaker is stationary, and the observer moves towards it with speed $\mathrm{u}$, the measured frequency is $f_{2}$. Then -
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The correct answer is:
$f_{1}>f_{2}$
$\mathrm{f}_{1}=\frac{\mathrm{f}_{0}[\mathrm{v}]}{\mathrm{v}-\mathrm{u}}$
$f_{2}=f_{0} \frac{[v+u]}{v}$
$f_{2}-f_{1}=f_{0}\left(\frac{u+v}{v}-\frac{v}{v-u}\right)$
$f_{2}-f_{1}=\frac{-u^{2} f_{0}}{(v)(v-u)}=-v e$
$\therefore f_{1}>f_{2}$
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