JEE Mains · Physics · STD 11 - 14. waves and sound
In a closed organ pipe, the frequency of fundamental note is \(30 \mathrm{~Hz}\). A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to \(110 \mathrm{~Hz}\). If the organ pipe has a cross-sectional area of \(2 \mathrm{~cm}^2\), the amount of water poured in the organ tube is _______ \(g.\) (Take speed of sound in air is \(330 \mathrm{~m} / \mathrm{s}\) )
- A \(400\)
- B \(200\)
- C \(600\)
- D \(800\)
Answer & Solution
Correct Answer
(A) \(400\)
Step-by-step Solution
Detailed explanation
\( \frac{V}{4 \ell_1}=30 \Rightarrow \ell_1=\frac{11}{4} \mathrm{~m} \) \( \frac{V}{4 \ell_2}=110 \Rightarrow \ell_2=\frac{3}{4} \mathrm{~m} \) \( \Delta \ell=2 \mathrm{~m},\) Change in volume \(=A \Delta \ell=400 \mathrm{~cm}^3\)…
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