JEE Main 2025 Apr 4 Shift 1, Physics Q16: Stationary Waves in Air Column

JEE Main2025Apr 4, Shift 1Physics

In an experiment with a closed organ pipe, it is filled with water by $(\frac{1}{5})^{th}$ of its volume. The frequency of the fundamental note will change by

A
25%
B
20%
C
$- 20%$
D
$- 25%$

Solution

For a closed organ pipe of air column length $L$ , the fundamental frequency is $f = \frac{v}{4 L}$ .

Initially the air column has length $L$ . After filling 1/5 of the pipe with water, the remaining air column is $\frac{4 L}{5}$ , so the new fundamental is $f^{'} = \frac{v}{4 \cdot ( 4 L /5 )} = \frac{5 v}{16 L}$ .

Percentage change $= \frac{f ^{'} - f}{f} \times 100 = (\frac{5/16}{1/4} - 1) \times 100 = (\frac{5}{4} - 1) \times 100 = 25%$ .

Frequency increases by 25%.

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JEE Main 2025 Apr 4 Shift 1, Physics Q16: Stationary Waves in Air Column

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