Question:
In an experiment with a closed organ pipe, it is filled with water by \(\left(\frac{1}{5}\right)^{th}\) of its volume. The frequency of the fundamental note will change by
In an experiment with a closed organ pipe, it is filled with water by \(\left(\frac{1}{5}\right)^{th}\) of its volume. The frequency of the fundamental note will change by
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Filling one fifth of the length with water shortens the air column to four fifths, and frequency is inversely proportional to length.
Updated On: Nov 12, 2026
- \(25\%\)
- \(20\%\)
- \(-20\%\)
- \(-25\%\)
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The Correct Option is A
Solution and Explanation
Step 1: Understanding the Concept
For a closed pipe the fundamental frequency is \(f=\dfrac{v}{4L}\), so \(f\propto\dfrac{1}{L}\).
Step 2: New length
Water fills \(\dfrac{1}{5}\) of the pipe, so the air column is \(L'=\dfrac{4L}{5}\).
Step 3: New frequency
\(f'=\dfrac{v}{4L'}=\dfrac{5}{4}f\).
Step 4: Percentage change
\(\dfrac{f'-f}{f}\times100=25\%\), option (A).
Final Answer:
The frequency increases by 25 percent. \[ \boxed{\text{(A) }25\%} \]
For a closed pipe the fundamental frequency is \(f=\dfrac{v}{4L}\), so \(f\propto\dfrac{1}{L}\).
Step 2: New length
Water fills \(\dfrac{1}{5}\) of the pipe, so the air column is \(L'=\dfrac{4L}{5}\).
Step 3: New frequency
\(f'=\dfrac{v}{4L'}=\dfrac{5}{4}f\).
Step 4: Percentage change
\(\dfrac{f'-f}{f}\times100=25\%\), option (A).
Final Answer:
The frequency increases by 25 percent. \[ \boxed{\text{(A) }25\%} \]
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