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NEET PHYSICSWAVE OPTICSMedium

Question

The 4th4^{\text{th}} overtone of a closed organ pipe is the same as that of the 3rd3^{\text{rd}} overtone of an open pipe. The ratio of the length of the closed pipe to the length of the open pipe is:

A

8:98:9

B

9:79:7

C

9:89:8

D

7:97:9

Step-by-Step Solution

For a closed organ pipe, the fundamental frequency is fc=v4Lcf_{c} = \frac{v}{4L_c}. The overtones are odd harmonics. The nthn^{\text{th}} overtone corresponds to the (2n+1)th(2n+1)^{\text{th}} harmonic. Frequency of the 4th4^{\text{th}} overtone of the closed pipe = (2×4+1)fc=9(v4Lc)(2 \times 4 + 1) f_c = 9 \left(\frac{v}{4L_c}\right).

For an open organ pipe, the fundamental frequency is fo=v2Lof_{o} = \frac{v}{2L_o}. The overtones are integer harmonics. The nthn^{\text{th}} overtone corresponds to the (n+1)th(n+1)^{\text{th}} harmonic. Frequency of the 3rd3^{\text{rd}} overtone of the open pipe = (3+1)fo=4(v2Lo)=2vLo(3 + 1) f_o = 4 \left(\frac{v}{2L_o}\right) = \frac{2v}{L_o}.

Given that these two frequencies are equal: 9v4Lc=2vLo\frac{9v}{4L_c} = \frac{2v}{L_o} LcLo=98\frac{L_c}{L_o} = \frac{9}{8}.

Therefore, the ratio of the length of the closed pipe to the length of the open pipe is 9:89:8.

Exam Context & Concepts Covered

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

PHYSICSWAVE OPTICStextthovertoneclosedtextrdovertone

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