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NEET PHYSICSWAVEMedium

Question

The fundamental frequency of a closed organ pipe of a length 20 cm20 \text{ cm} is equal to the second overtone of an organ pipe open at both ends. The length of the organ pipe open at both ends will be:

A

80 cm80 \text{ cm}

B

100 cm100 \text{ cm}

C

120 cm120 \text{ cm}

D

140 cm140 \text{ cm}

Step-by-Step Solution

Let the length of the closed organ pipe be Lc=20 cmL_c = 20 \text{ cm}. The fundamental frequency of a closed organ pipe is given by fc=v4Lcf_c = \frac{v}{4L_c}. Let the length of the open organ pipe be LoL_o. The second overtone of an open organ pipe corresponds to the third harmonic. Its frequency is given by fo=3v2Lof_o = \frac{3v}{2L_o}. According to the question, fc=fof_c = f_o: v4Lc=3v2Lo\frac{v}{4L_c} = \frac{3v}{2L_o} Rearranging to solve for LoL_o: Lo=3v×4Lc2v=6LcL_o = \frac{3v \times 4L_c}{2v} = 6L_c Given Lc=20 cmL_c = 20 \text{ cm}: Lo=6×20 cm=120 cmL_o = 6 \times 20 \text{ cm} = 120 \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.

PHYSICSWAVEfundamentalfrequencyclosedlengthsecond

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