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RAJASTHAN PMT Rajasthan - PMT Solved Paper-2001

  • question_answer
    Minimum resonating length of a closed organ pipe is 50 cm, when it is vibrated by same frequency, then next resonating length will be:

    A)  250 cm                                

    B)  200 cm

    C)  150 cm  

    D)         100 cm

    Correct Answer: C

    Solution :

                    For closed pipe\[{{l}_{1}}:{{l}_{2}}=1:3\] \[\therefore \]  \[\frac{50}{{{l}_{2}}}=\frac{1}{3}\] \[\therefore \]  \[{{l}_{2}}=150\,cm\]


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