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JEE Main & Advanced Sample Paper JEE Main - Mock Test - 12

  • question_answer
    A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is \[340\,\text{m}{{\text{s}}^{-1}}\].

    A) 4                     

    B) 12   

    C) 8                     

    D) 6

    Correct Answer: D

    Solution :

    [d]: Here, \[l=85\text{ }cm=0.85m,v=340\text{ }m\text{ }{{s}^{-}}^{1}\] Pipe is closed from one end so it behaves as a closed organ pipe. Frequencies in the closed organ pipe is given by, \[\upsilon =\frac{(2n-1)v}{4l}\]where, n = 1, 2, 3, 4,........... According to question,\[\upsilon <1250Hz\] \[\left( \frac{2n-1}{4l} \right)v<1250\Rightarrow \frac{(2n-1)\times 340}{4\times 0.85}<1250\] \[\Rightarrow \]\[(2n-1)<12.5\] Possible value of n = 1, 2, 3, 4, 5, 6 So, number of possible natural frequencies lie below 1250 Hz is 6.


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