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  3. RAJASTHAN PMT

RAJASTHAN PMT Rajasthan - PMT Solved Paper-2002

  • question_answer
    26 tuning fork are arranged in a line having beat frequency of 4 between two successive tuning forks. If frequency of last tuning fork is 3 times that of 1st tuning fork then find the frequency of 1st tuning fork:

    A)  7.5 Hz                  

    B)         50 Hz

    C)  100 Hz

    D)         125 Hz

    Correct Answer: B

    Solution :

    Beat frequency of two successive tuning forks is 4 \[\therefore \] difference between the frequencies = 4 From formula of A.P. \[\begin{align}   & I=a+(n-1)\,d \\  & 3f=f+(26-1)\times 4 \\  & 2f=100 \\  & f=50Hz \\ \end{align}\]


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