A pipe open at both ends produces a fundamental frequency n₁. When the pipe is kept with 3/4th of its length in water, it produces a note of fundamental frequency n₂. The ratio of (n₁)/(n₂) is (a) 4/3 (b) 3/4 (c) 2 (d) 1/2

Answer: (d). Open pipe: n₁ = v/2L. With 3/4 in water it becomes a closed pipe of length L/4: n₂ = (v)/(4(L/4)) = v/L. (n₁)/(n₂) = (v/2L)/(v/L) = 1/2 So the answer is (d) 1/2.

A pipe open at both ends produces a fundamental frequency n₁. When… | MHT CET 2025 (23 Apr, Shift 1)

Q31MHT CET 2025 · 23 APR · SHIFT 1Easy

Organ Pipes & Resonance Tube

A pipe open at both ends produces a fundamental frequency $n_1$ . When the pipe is kept with ${\frac{3}{4}}^{\text{th}}$ of its length in water, it produces a note of fundamental frequency $n_2$ . The ratio of $\frac{n_1}{n_2}$ is

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Answer: (d)

Open pipe: $n_1 = \frac{v}{2L}$ .
With $\frac{3}{4}$ in water it becomes a closed pipe of length $\frac{L}{4}$ : $n_2 = \frac{v}{4(L/4)} = \frac{v}{L}$ .
$\frac{n_1}{n_2} = \frac{v/2L}{v/L} = \frac{1}{2}$
So the answer is (d) $\frac{1}{2}$ .

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Organ Pipes & Resonance Tube question from MHT CET 2025 (23 Apr, Shift 1)

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