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NEET generalGeneralMedium

Question

The fundamental frequency in an open organ pipe is equal to the third harmonic of a closed organ pipe. If the length of the closed organ pipe is 20 cm, the length of the open organ pipe is

A

12.5 cm

B

8 cm

C

13.2 cm

D

16 cm

Step-by-Step Solution

For a closed organ pipe, the third harmonic frequency is fc=3v4lf_c = \frac{3v}{4l}. For an open organ pipe, the fundamental frequency is fo=v2lf_o = \frac{v}{2l'}. Given fo=fcf_o = f_c, so v2l=3v4l\frac{v}{2l'} = \frac{3v}{4l}. Thus, l=4l3×2=2l3=2×203=13.33 cml' = \frac{4l}{3 \times 2} = \frac{2l}{3} = \frac{2 \times 20}{3} = 13.33 \text{ cm}. Note: The provided answer key in the image says (3) 13.2 cm, but the calculation yields 13.33 cm.

Exam Context & Concepts Covered

This question aligns with the NEET general syllabus, specifically targeting concepts from General. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

generalfundamentalfrequencyharmonicclosedlength

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