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An observer moves towards a stationary source of sound, with a speed of one-fifth of the speed of sound. The apparent increase in the frequency heard by the observer is
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$20 \%$
From the Doppler's effect
$$
f^{\prime}=f_0\left(\frac{v_0 \pm v_0}{v \pm v_s}\right)
$$
$\mathrm{v}_{\mathrm{s}}=$ velocity of source
$\mathrm{v}_0=$ velocity of observer
$\mathrm{y}=$ velocity of sound
Here $\mathrm{v}_{\mathrm{s}}=0$
$\begin{aligned} & f^{\prime}=f_o\left(\frac{v+v_0}{v}\right) \Rightarrow \frac{f^{\prime}}{f_0}=\frac{v+\frac{v}{5}}{v} \\ & =\frac{6}{5} \\ & \frac{f^{\prime}-f_o}{f_o} \times 100 \%=\frac{6-5}{5}=\frac{1}{5} \times 100 \% \\ & =20 \%\end{aligned}$
$$
f^{\prime}=f_0\left(\frac{v_0 \pm v_0}{v \pm v_s}\right)
$$
$\mathrm{v}_{\mathrm{s}}=$ velocity of source
$\mathrm{v}_0=$ velocity of observer
$\mathrm{y}=$ velocity of sound
Here $\mathrm{v}_{\mathrm{s}}=0$
$\begin{aligned} & f^{\prime}=f_o\left(\frac{v+v_0}{v}\right) \Rightarrow \frac{f^{\prime}}{f_0}=\frac{v+\frac{v}{5}}{v} \\ & =\frac{6}{5} \\ & \frac{f^{\prime}-f_o}{f_o} \times 100 \%=\frac{6-5}{5}=\frac{1}{5} \times 100 \% \\ & =20 \%\end{aligned}$
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