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JEE Main & Advanced Physics Wave Mechanics Question Bank Organ Pipe (Vibration of Air Column)

  • question_answer
    In a resonance pipe the first and second resonances are obtained at depths 22.7 cm and 70.2 cm respectively. What will be the end correction                                    [J & K CET 2005]

    A)            1.05 cm                                   

    B)            115.5 cm

    C)            92.5 cm                                   

    D)            113.5 cm

    Correct Answer: A

    Solution :

                       For end correction x,  \[\frac{{{l}_{2}}+x}{{{l}_{1}}+x}=\frac{3\lambda /4}{\lambda /4}=3\]            \[x=\frac{{{l}_{2}}-3{{l}_{1}}}{2}=\frac{70.2-3\times 22.7}{2}=1.05\,cm\]


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