1.

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

Answer» Correct Answer - A
When source is approaching the observer, the frequency heard
`n_a=((v)/(v-v_S))xxn=((340)/(340-20))xx1000=1063Hz`
When source is receding the frequency heard
`n_r=((v)/(v+v_S))xxn=(340)/(340-20)xx1000=944`
`impliesn_a:n_r=9:8`
Short tricks : `(n_a)/(n_r)=(v+v_S)/(v-v_S)=(340+20)/(340-20)=(9)/(8)`
(speed of sound `v=340m//s`).


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