2. Sample Papers
  3. NEET

NEET Sample Paper NEET Sample Test Paper-39

  • question_answer
    The fundamental frequency of a closed organ pipe of length 20 cm is equal to the second overtone of an organ pipe open at both the ends. The length of organ pipe open at both the ends is

    A)  80 cm       

    B)  100 cm

    C)  120 cm      

    D)  140 cm

    Correct Answer: C

    Solution :

     Fundamental frequency of closed organ pipe \[{{f}_{c}}=\frac{v}{4{{l}_{c}}}\] Fundamental frequency of open organ pipe \[{{f}_{0}}=\frac{v}{2{{l}_{0}}}\] Second overtone frequency of open organ pipe, \[=\frac{3v}{2{{l}_{0}}}\] From question, \[\frac{v}{4{{l}_{c}}}=\frac{3v}{2{{l}_{0}}}\] \[{{l}_{0}}=6{{l}_{c}}=6\times 20=120\,cm\]


You need to login to perform this action.
You will be redirected in 3 sec spinner