Q124 / 150 · MHT CET 2023 (11 May Shift 1) · Physics

The second overtone of an open pipe has the same frequency as the first overtone of a closed pipe of length ' $L$ '. The length of the open pipe will be

  • A
    $\frac{L}{2}$
  • B
    $\mathrm{L}$
  • Correct answer
    C
    $\mathrm{2L}$
  • D
    $\mathrm{4L}$

Solution

The length of closed pipe is denoted using $\mathrm{L}$. Let $l$ be the length of open pipe and $\mathrm{v}$ be the velocity. Frequency of second overtone of an open organ pipe is $\mathrm{n}_{\mathrm{o}}=\frac{3 \mathrm{v}}{2 l}$ Frequency of first overtone of a closed pipe is $\mathrm{n}_{\mathrm{c}}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}$ Given: $\mathrm{n}_0=\mathrm{n}_{\mathrm{c}}$ $\frac{3 \mathrm{v}}{2 l}=\frac{3 \mathrm{v}}{4 \mathrm{~L}}$ $\mathrm{L}=\frac{l}{2}$ $\therefore \quad l=2 \mathrm{~L}$

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

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