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JEE Main & Advanced Physics Wave Mechanics Question Bank Mock Test - Waves and Acoustics

  • question_answer
    An open pipe is in resonance in its \[{{2}^{nd}}\] harmonic with tuning fork of frequency\[{{f}_{1}}\]. Now it is closed at one end. If the frequency of the tuning fork is increased slowly from \[{{f}_{1}}\] then again a resonance is obtained with a frequency\[{{f}_{2}}\]. If in this case the pipe vibrates \[{{n}^{th}}\] harmonics then

    A) \[n=3,\,\,{{f}_{2}}=\frac{3}{4}{{f}_{1}}\]      

    B) \[n=3,\,\,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\]

    C) \[n=5,\,\,{{f}_{2}}=\frac{5}{4}{{f}_{1}}\]      

    D) \[n=5,\,\,{{f}_{2}}=\frac{3}{4}{{f}_{1}}\]

    Correct Answer: C

    Solution :

    [c] Open pipe resonance frequency \[{{f}_{1}}=\frac{2v}{2L}\] Closed pipe resonance frequency\[{{f}_{2}}=\frac{nv}{4L}\] \[{{f}_{2}}=\frac{n}{4}{{f}_{1}}\]    (Where n is odd and\[{{f}_{2}}>{{f}_{1}}\]) \[\therefore n=5\]


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