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For a closed organ pipe, the first minimum resonating length is L₁ = λ/4 = 50 cm. Next or second resonating length, L₂ = 3λ/4 = 150 cm. If an open pipe is half submerged in water, it will become a closed organ pipe of length half that of an open pipe. Its fundamental frequency will become n' = V/(ℓ/2) = V/2ℓ equal to that of an open pipe.

1. L₁ = λ/4 = 50 cm
2. L₂ = 3λ/4 = 150 cm
3. n' = V/(ℓ/2) = V/2ℓ
4. Fundamental frequency will equal that of an open pipe

Correct answer: Fundamental frequency will equal that of an open pipe

Solution

The fundamental frequency of a closed organ pipe is equal to that of an open pipe when the effective length of the closed pipe is half the length of the open pipe. This matches the condition described in the question.

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