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BHU PMT BHU PMT (Mains) Solved Paper-2007

  • question_answer
    Two tuning forks when sounded together give 8 beat/s. When A is sounded with air column of length 37.5 cm closed at one end resonance occurs in its fundamental mode. B gives resonance with air column of length 38.5 cm and closed at one end in its fundamental mode. The frequencies of tuning forks are

    A)  300 Hz, 300 Hz  

    B)  300 Hz, 308 Hz

    C)  308 Hz, 308 Hz  

    D)  308 Hz, 300 Hz

    Correct Answer: D

    Solution :

                     Given \[{{f}_{A}}-{{f}_{B}}=8\]                                ?. (i) \[{{f}_{A}}=\frac{v}{4l}=\frac{v}{4\times 37.5\times {{10}^{-2}}}\] and         \[{{f}_{B}}=\frac{v}{4\times 38.5\times {{10}^{-2}}}\] So,          \[\frac{{{f}_{A}}}{{{f}_{B}}}=\frac{38.5}{37.5}\] \[\Rightarrow \]               \[{{f}_{A}}=\frac{385{{f}_{B}}}{375}\]                     ??. (ii)  Putting value of f^ in Eq. (i), we get \[\frac{385{{f}_{B}}}{375}-{{f}_{B}}=8\] \[\Rightarrow \]               \[10{{f}_{B}}=8\times 375\] \[\therefore \]  \[{{f}_{B}}=\frac{8\times 375}{10}=300\,Hz\] Hence, from Eq. (i), we get \[{{f}_{A}}=8+300=308\,Hz\]


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