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NEET PHYSICSWAVEMedium

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 cm20 \text{ cm}, the length of the open organ pipe is:

A

13.2 cm13.2 \text{ cm}

B

8 cm8 \text{ cm}

C

12.5 cm12.5 \text{ cm}

D

16 cm16 \text{ cm}

Step-by-Step Solution

Let the length of the open organ pipe be LoL_o and the length of the closed organ pipe be LcL_c. The fundamental frequency of an open organ pipe is given by fo=v2Lof_o = \frac{v}{2L_o}. The frequency of the third harmonic (first overtone) of a closed organ pipe is given by fc=3v4Lcf_c = \frac{3v}{4L_c}. According to the question, fo=fcf_o = f_c: v2Lo=3v4Lc\frac{v}{2L_o} = \frac{3v}{4L_c} Rearranging to solve for LoL_o: Lo=4Lc6=2Lc3L_o = \frac{4L_c}{6} = \frac{2L_c}{3} Given Lc=20 cmL_c = 20 \text{ cm}, Lo=2×203=403=13.33 cmL_o = \frac{2 \times 20}{3} = \frac{40}{3} = 13.33 \text{ cm} The calculated length is 13.33 cm13.33 \text{ cm}, and the closest available option provided in the question is 13.2 cm13.2 \text{ cm}.

Exam Context & Concepts Covered

This question aligns with the NEET PHYSICS syllabus, specifically targeting concepts from WAVE. 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.

PHYSICSWAVEfundamentalfrequencyharmonicclosedlength

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