Q118 / 180 · NEET 2015 (Phase 1) · Physics

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 the organ pipe open at both the ends is

  • A
    80 cm
  • B
    100 cm
  • Correct answer
    C
    120 cm
  • D
    140 cm

Solution

For closed organ pipe, \(n_c=\frac{V}{4 l}\) For open organ pipe, \(n_0=\frac{V}{2 l^{\prime}}\) For the second overtone of an open organ pipe, \(\begin{aligned} & n^{\prime}=3 n_0=\frac{3 V}{2 l^{\prime \prime}} \\ & n_c=n^{\prime} \\ & \frac{V}{4 l}=\frac{3 V}{2 l^{\prime}} \\ & l^{\prime}=6 l=6 \times 20 \\ & l^{\prime}=120 \mathrm{~cm} \end{aligned}\)

Hence, the correct answer is option: 'C'.

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