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NEET Sample Paper NEET Sample Test Paper-33

  • question_answer
    A train approaches a stationary observer, the velocity of train being 1/20 of the velocity of sound. A sharp blast is blowen with the whistle of the engine at equal intervals of a second. The interval between the successive blasts as. heard by the observer:

    A)  \[\frac{1}{20}S\]            

    B)  \[1/20\,min\]

    C)  \[19/20\text{ }S\]           

    D)  \[19/20\,min\]

    Correct Answer: C

    Solution :

    According to Doppler effect in sound the app change in frequency of the source due to a relative motion between source and observer is: \[n'=n\left( \frac{V-{{V}_{0}}}{V-{{V}_{S}}} \right)\] Where V is velocity of sound, \[{{V}_{0}}\] the velocity of observer and \[{{V}_{S}}\] the velocity of source. Here,       \[{{V}_{0}}=0\] (observer stationary) \[{{V}_{S}}=\frac{V}{20}\] \[n=1Hz.\] (as blast in blown at an interval of Is) \[n'=\frac{V}{V-\frac{V}{20}}\times 1=\frac{20}{19}Hz.\] Observed time interval between two successive blasts \[=\frac{19}{20}S.\]


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