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A source producing sound of frequency $170 \mathrm{~Hz}$ is approaching a stationary observer with a velocity $17 \mathrm{~ms}^{-1}$. The apparent change in the wavelength of sound heard by the observer is (speed of sound in air $=340 \mathrm{~ms}^{-1}$ )
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The correct answer is:
$0.1 \mathrm{~m}$
$$
\begin{aligned}
& \lambda=\frac{v}{n}=\frac{340}{170}=2 m, \quad n^{\prime}=\frac{340}{340-17} \times 170 \\
& n^{\prime}=178.9 \mathrm{~Hz} \\
& \text { Now } \lambda^{\prime}=\frac{v}{n^{\prime}}=\frac{340}{178.9}=1.9 \\
& \Rightarrow \lambda-\lambda^{\prime}=2-1.9=0.1 \mathrm{~m}
\end{aligned}
$$
\begin{aligned}
& \lambda=\frac{v}{n}=\frac{340}{170}=2 m, \quad n^{\prime}=\frac{340}{340-17} \times 170 \\
& n^{\prime}=178.9 \mathrm{~Hz} \\
& \text { Now } \lambda^{\prime}=\frac{v}{n^{\prime}}=\frac{340}{178.9}=1.9 \\
& \Rightarrow \lambda-\lambda^{\prime}=2-1.9=0.1 \mathrm{~m}
\end{aligned}
$$
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