NEET Sample Paper NEET Sample Test Paper-14

  • 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)  8:9

    C)  1:1                             

    D)  9:10

    Correct Answer: A

    Solution :

    \[f={{f}_{o}}\left( \frac{v+{{v}_{o}}}{v-{{v}_{s}}} \right)\] 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].


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