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NEET Sample Paper NEET Sample Test Paper-1

  • question_answer
    If length of a closed organ pipe is \m and velocity of sound is\[330\text{ }mis\]. Then, the frequency of second note will be:

    A)  \[2\times \frac{4}{330}\,Hz\]                

    B)  \[2\times \frac{330}{4}\,Hz\]

    C)  \[3\times \frac{330}{4}\,Hz\]                    

    D)  \[4\times \frac{330}{4}\,Hz\]

    Correct Answer: C

    Solution :

    Given length of closed organ pipe \[=1\,m\] The frequency of closed organ pipe is             \[\frac{V}{4l},\frac{3V}{4l},\frac{5V}{4l}....\] Therefore frequency of second note is             \[\frac{3V}{4l}=3\times \frac{330}{4\times 1}=\frac{3\times 330}{4}\,Hz.\]


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