Q140 / 160 · TS EAMCET 2023 (13 May Shift 1) · Physics

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

A
$16.67 \%$
B
$2 \%$
C
$25 \%$
Correct answer
D
$20 \%$

Solution

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 $\mathrm{v}_{\mathrm{s}}=0$ $\begin{aligned} & f^{\prime}=f_o\left(\frac{v+v_0}{v}\right) \Rightarrow \frac{f^{\prime}}{f_0}=\frac{v+\frac{v}{5}}{v} \\ & =\frac{6}{5} \\ & \frac{f^{\prime}-f_o}{f_o} \times 100 \%=\frac{6-5}{5}=\frac{1}{5} \times 100 \% \\ & =20 \%\end{aligned}$

Hence, the correct answer is option: 'D'.

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