16:I[8216,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"BackButton"] 17:I[1553,["972","static/chunks/972-b28da0de92681928.js","47","static/chunks/47-a7f06e9eda1ac756.js","115","static/chunks/app/%5Bexam%5D/questions/%5Byear%5D/%5Bsubject%5D/%5Bchapter%5D/%5BquestionId%5D/page-7b6f0a12b9cc0206.js"],"QuestionViewer"] 15:T4c1,{"@context":"https://schema.org","@type":"Quiz","name":"Comedk 2024 Waves PYQ with Solution","url":"https://pyq-test-series-platform-alsjdcxu2-sameers-projects-41c5a71e.vercel.app/comedk/questions/2024/physics/waves/lx0kccge","educationalLevel":"Entrance exam preparation","about":"Comedk / Physics / Waves","hasPart":{"@type":"Question","name":"Comedk 2024 Waves PYQ with Solution","text":"The number of possible natural oscillations of air column in a pipe closed at one end of length 85 \\mathrm{~cm} whose frequencies lie below 1250 \\mathrm{~Hz} are (velocity of sound =340 \\mathrm{~ms}^{-1})","eduQuestionType":"mcq","acceptedAnswer":{"@type":"Answer","text":"A"},"comment":{"@type":"Comment","text":"For a pipe closed at one end, the fundamental frequency is given by: f_1 = \\frac{v}{4L} where: f_1 is the fundamental frequency v is the speed of sound L is the length of the pipe The possible frequencies for a pipe closed at one end are odd multiples of the fundamental frequency: f_n = (2n-1)f_1 = (2n-1)\\frac{v}{4L} where n is an integer (1, 2, 3, ...). We need to find the number of possible frequencies below 1250 \\mathrm{~Hz}. Let's plug in the given values and solve for n: 1250 \\mathrm{~Hz}..."}}}18:T458,

For a pipe closed at one end, the fundamental frequency is given by:

$$f_1 = \frac{v}{4L}$$

where:

$$f_1$$ is the fundamental frequency

$$v$$ is the speed of sound

$$L$$ is the length of the pipe

The possible frequencies for a pipe closed at one end are

Waves

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