A detector is released from rest over a source of sound of frequency $f = 10^3 \,\,Hz$. The frequency observed by the detector at time $t$ is plotted in the graph. The speed of sound in air is $(g = 10 \,\,m/s^2)$ ... $m/s$
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$f=f_{o}\left(\frac{\nu+\nu_{o}}{\nu}\right)$
$=10^{3}\left(1+\frac{10 t}{\nu}\right)\left(\text { as } \nu_{o}=g t\right)$
Hence, $f$ versus $t$ graph is straight line of slope $\frac{10^{4}}{\nu}$
$\therefore \frac{10^{4}}{\nu}=$ slope $=\frac{100}{3}$
$\therefore \nu=300 \mathrm{m} / \mathrm{s}$
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