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NEET Sample Paper NEET Sample Test Paper-89

  • question_answer
    A pipe closed at one end produces a fundamental note of 412 Hz. It is cut into two pieces of equal length. The fundamental frequencies produced by the two pieces are

    A) 206 Hz, 412 Hz

    B)        824 Hz, 1648 Hz

    C) 412 Hz, 824 Hz

    D)        206 Hz, 824 Hz

    Correct Answer: B

    Solution :

    [b] \[\frac{v}{4\ell }=412,\] when cut into two equal pieces, frequency of closed pipe of half the length \[=\frac{v}{4(\ell /2)}=\frac{2v}{4\ell }=2\times 412=824\,\,Hz\] frequency of open pipe of half the length \[=\frac{v}{2(\ell /2)}=4\times 412=1648\,\,Hz\]


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