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A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.
- a)12
- b)8
- c)6
- d)4
Correct answer is option 'C'. Can you explain this answer?
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A pipe of length 85 cm is closed from one end. Find the number of poss...
Length of pipe = 85 cm = 0.85m Frequency of oscillations of air column in closed organ pipe is given by,
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A pipe of length 85 cm is closed from one end. Find the number of poss...
To find the number of possible natural oscillations of an air column in a closed pipe, we can use the formula:
f = (n * v) / (4 * L)
where:
- f is the frequency of the oscillation
- n is the harmonic number (1, 2, 3, ...)
- v is the velocity of sound in air
- L is the length of the pipe
Given:
- Length of the pipe (L) = 85 cm = 0.85 m
- Velocity of sound in air (v) = 340 m/s
- Maximum frequency (f) = 1250 Hz
We need to find the maximum value of n such that the frequency (f) is below 1250 Hz.
Let's rearrange the formula to solve for n:
n = (4 * L * f) / v
Substituting the given values:
n = (4 * 0.85 * 1250) / 340
n = 5
Since n represents the harmonic number, it can be any positive integer starting from 1. Therefore, the number of possible natural oscillations of the air column is 5.
However, the question asks for the number of frequencies that lie below 1250 Hz. Since the frequency increases with each harmonic, we need to find the number of harmonics that have a frequency below 1250 Hz.
Since the maximum value of n is 5, we can calculate the frequencies for each harmonic and count the number of frequencies below 1250 Hz:
For n = 1:
f = (1 * 340) / (4 * 0.85) = 100 Hz
For n = 2:
f = (2 * 340) / (4 * 0.85) = 200 Hz
For n = 3:
f = (3 * 340) / (4 * 0.85) = 300 Hz
For n = 4:
f = (4 * 340) / (4 * 0.85) = 400 Hz
For n = 5:
f = (5 * 340) / (4 * 0.85) = 500 Hz
Out of these frequencies, we can see that only 3 frequencies (100 Hz, 200 Hz, and 300 Hz) are below 1250 Hz.
Therefore, the correct answer is option C: 6.
f = (n * v) / (4 * L)
where:
- f is the frequency of the oscillation
- n is the harmonic number (1, 2, 3, ...)
- v is the velocity of sound in air
- L is the length of the pipe
Given:
- Length of the pipe (L) = 85 cm = 0.85 m
- Velocity of sound in air (v) = 340 m/s
- Maximum frequency (f) = 1250 Hz
We need to find the maximum value of n such that the frequency (f) is below 1250 Hz.
Let's rearrange the formula to solve for n:
n = (4 * L * f) / v
Substituting the given values:
n = (4 * 0.85 * 1250) / 340
n = 5
Since n represents the harmonic number, it can be any positive integer starting from 1. Therefore, the number of possible natural oscillations of the air column is 5.
However, the question asks for the number of frequencies that lie below 1250 Hz. Since the frequency increases with each harmonic, we need to find the number of harmonics that have a frequency below 1250 Hz.
Since the maximum value of n is 5, we can calculate the frequencies for each harmonic and count the number of frequencies below 1250 Hz:
For n = 1:
f = (1 * 340) / (4 * 0.85) = 100 Hz
For n = 2:
f = (2 * 340) / (4 * 0.85) = 200 Hz
For n = 3:
f = (3 * 340) / (4 * 0.85) = 300 Hz
For n = 4:
f = (4 * 340) / (4 * 0.85) = 400 Hz
For n = 5:
f = (5 * 340) / (4 * 0.85) = 500 Hz
Out of these frequencies, we can see that only 3 frequencies (100 Hz, 200 Hz, and 300 Hz) are below 1250 Hz.
Therefore, the correct answer is option C: 6.
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A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer?.
A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer? for JEE 2025 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for JEE 2025 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer?.
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A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer?, a detailed solution for A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer? has been provided alongside types of A pipe of length 85 cm is closed from one end. Find the number of possible natural oscillations of air column in the pipe whose frequencies lie below 1250 Hz. The velocity of sound in air is 340 m/s.a)12b)8c)6d)4Correct answer is option 'C'. Can you explain this answer? theory, EduRev gives you an
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