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JEE Main & Advanced JEE Main Paper (Held On 26 May 2012)

  • question_answer
    An air column in a pipe, which is closed at one end, will be in resonance with a vibrating tuning fork of frequency 264 Hz if the length of the column in cm is (velocity of sound = 330 m/s)   JEE Main Online Paper (Held On 26-May-2012)  

    A) 125.00                                  

    B)                        93.75

    C)                         62.50                                   

    D)                        187.50

    Correct Answer: B

    Solution :

                    Given : Frequency of tuning fork, \[n=264Hz\] Length of column L= ? For closed organ pipe\[n=\frac{v}{4l}\] \[\Rightarrow \]\[l=\frac{v}{4n}=\frac{330}{4\times 264}=0.125\]or\[l=0.3125\times 100=31.25cm\]In case of closed organ pipe only odd harmonics are possible. Therefore value of \[l\] will be \[(2n-1)l\] Hence option (b) i.e. \[3\times 31.25=93.75cm\]is correct.


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