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In case of closed organ pipe frequency, fn = (2n + 1) v / 4l for n = 0, f0 = 100 Hz, n = 1, f1 = 300 Hz, n = 2, f2 = 500 Hz, n = 3, f3 = 700 Hz, n = 4, f4 = 900 Hz, n = 5, f5 = 1100 Hz, n = 6, f6 = 1300 Hz. Hence possible natural oscillation whose frequencies < 1250 Hz = 6(n = 0, 1, 2, 3, 4, 5).

1. n = 0, f0 = 100 Hz
2. n = 1, f1 = 300 Hz
3. n = 2, f2 = 500 Hz
4. n = 3, f3 = 700 Hz

Correct answer: n = 3, f3 = 700 Hz

Solution

The question asks for the possible natural oscillation frequencies less than 1250 Hz. For n = 3, f3 = 700 Hz, which satisfies the condition. Thus, option D is correct.

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