An observer moves towards a stationary source of sound with a velocity equal to one-fifth of the velocity of sound. The percentage change in the frequency will be $\dots \;$%
JEE MAIN 2022, Medium
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$f_{0}=\left(\frac{v+v_{0}}{v}\right) f_{s}$
$f_{0}=\left(\frac{v+\frac{v}{5}}{v}\right) f_{s}$
$f_{0}=\frac{6}{5} f_{s}$
$\%$ change $=\frac{f_{0}-f_{s}}{f_{s}} \times 100$
$=\frac{1}{5} \times 100=20 \%$
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