BVP Medical BVP Medical Solved Paper-2007

  • question_answer
    A tuning fork of frequency 392 Hz, resonates with 50 cm length of a string under tension (T). If length of the string is decreased by 2%, keeping the tension constant, the number of beats heard when the string and the tuning fork made to vibrate simultaneously is

    A)                 4                                             

    B)                 6

    C)                 8                                             

    D)                 12

    Correct Answer: C

    Solution :

                    Key Idea: According to law of length \[\left[ {{M}^{0}}L{{T}^{-1}} \right]\] The frequency of tuning fork is given by                 \[\left[ {{M}^{0}}{{L}^{-1}}{{T}^{0}} \right]\] where I is length of string, T is tension and m is mass per unit length of string. Given, \[\left[ {{M}^{0}}LT \right]\]                 \[\frac{3}{2}m{{R}^{2}}\] \[\frac{2}{3}m{{R}^{2}}\]              \[\frac{5}{4}m{{R}^{2}}\] Also number of beats per second (beat frequency) \[\frac{4}{5}m{{R}^{2}}\] difference of the frequencies of sound-sources. \[4.58\times {{10}^{-6}}J/{{m}^{3}}\]


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