• question_answer A source of sound is moving with constant velocity of 20 m/s emitting a note of frequency 1000 Hz. The ratio of frequencies observed by a stationary observer while the source is approaching him and after it crosses him will be:(Speed of sound v = 340 m/s) A)  9:8                             B)  8:9C)  1:1                             D)  9:10

$f={{f}_{o}}\left( \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right)$ Apparent frequency when the source is approaching the observer$={{f}_{1}}$ $\therefore$${{f}_{1}}={{f}_{0}}\left[ \frac{v+o}{v-20} \right]$ Apparent frequency when the source recedes from the observer$={{f}_{2}}$ $\therefore$${{f}_{2}}={{f}_{o}}\left[ \frac{v}{v-(-20)} \right]={{f}_{o}}\left[ \frac{v}{v+20} \right].$ $\therefore$    $\frac{{{f}_{1}}}{{{f}_{2}}}=\frac{v+20}{v-20}=\frac{340+20}{340-20}=\frac{9}{8}$ Hence, the correction option is [a].