Uttarakhand PMT Uttarakhand PMT Solved Paper-2006

  • question_answer
    A closed argon pipe and an open argon pipe are tuned to the same fundamental frequency. What is the ratio of their lengths?

    A)  1 : 2            

    B)  2 : 1

    C)  2 : 3            

    D)  4 : 3

    Correct Answer: A

    Solution :

     Avoiding end correction, the length of closed organ pipe is \[{{l}_{1}}=\frac{{{\lambda }_{1}}}{4}\]or\[{{l}_{2}}=4{{l}_{1}}\] The length of open organ pipe is \[{{l}_{2}}=\frac{{{\lambda }_{2}}}{2}\]or\[{{\lambda }_{2}}=2{{l}_{2}}\] Here,      \[{{n}_{1}}={{n}_{2}}\] So, \[\frac{v}{{{\lambda }_{1}}}=\frac{v}{{{\lambda }_{2}}}\]or \[\frac{v}{4{{l}_{1}}}=\frac{v}{4{{l}_{2}}}\] Therefore, \[{{l}_{1}}:{{l}_{2}}=1:2\]


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