CET Karnataka Medical CET - Karnataka Medical Solved Paper-2002

  • question_answer
    A racing car moving towards a cliff, sounds its horn. The driver observes that the sound reflected from the cliff has a pitch one octave higher than the actual sound of the horn. If v is the velocity of sound, then the velocity of the car is:

    A)  \[\upsilon /2\]

    B)  \[\upsilon /\sqrt{2}\]

    C)   \[\upsilon /4\]

    D)  \[\upsilon /3\]

    Correct Answer: D

    Solution :

     Let f be the frequency of horn emitted by car. f = observed frequency \[\because \] pitch \[\propto f\] One octave = frequency of instrument \[\therefore \] \[f=\frac{\upsilon -{{\upsilon }_{O}}}{\upsilon +{{\upsilon }_{S}}}f\] \[{{\upsilon }_{O}}=\] velocity of observer let it be u. \[{{\upsilon }_{S}}=\] velocity of source because here source and observer (car) both are same and moving with same velocity. \[\therefore \]\[{{\upsilon }_{S}}={{\upsilon }_{O}}=u\]v = velocity of sound. AB = source and sound are in same direction. Hence \[(\upsilon -u)=\] relative velocity. BA = Observer and sound are in opposite direction hence \[\upsilon +u=\]relative velocity \[\therefore \] \[f=\frac{\upsilon -u}{\upsilon +u}f\] \[2f=\frac{\upsilon -u}{\upsilon +u}f\] \[2\upsilon +2u=\upsilon -u\] \[\upsilon =3u\] \[u=\upsilon /3.\]


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