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RAJASTHAN ­ PET Rajasthan PET Solved Paper-2005

  • question_answer
    Two steady sound sources having same frequency are placed at a distance of 1 m. An observer is moving at a distance of 10 m along the line joining there source. If the observed distance between first two successive maxima is 0.11 m, then frequency of sources is (velocity of sound v = 330 m/s.)           

    A)  1000 Hz         

    B)  3000 Hz       

    C)  3500 Hz         

    D)  4000 Hz         

    Correct Answer: B

    Solution :

     In second wave, the distance between maxima is called wavelength\[(\lambda )\] \[\lambda =0.11\,m\] The frequency of each source \[n=\frac{v}{\lambda }\] \[=\frac{330}{0.11}\] \[=3000\,Hz\]


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