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MHT CET 2025 (19 Apr, Shift 1) · Organ Pipes & Resonance Tube

Organ Pipes & Resonance Tube question from MHT CET 2025 (19 Apr, Shift 1)

Q41MHT CET 2025 · 19 APR · SHIFT 1Medium

Organ Pipes & Resonance Tube

The fundamental frequency of a closed pipe of length $L$ is equal to the second overtone of a pipe open at both the ends of length (XL). The value of X is (Neglect end correction)

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Answer: (c)

Closed pipe fundamental: $n_c = \frac{v}{4L}$ .
Open pipe second overtone (3rd harmonic): $n_o = \frac{3v}{2(XL)}$ .
$\frac{v}{4L} = \frac{3v}{2XL} \Rightarrow 2XL = 12L \Rightarrow X = 6$
So the answer is (c) 6.

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