JEE Mains · Physics · STD 11 - Waves and sound
- JEE Mains 2024
- Medium
- Numerical
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 \(330 \mathrm{~m} / \mathrm{s}\) )
Pick an option to check your answer.
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 \(=A \Delta \ell=400 \mathrm{~cm}^3\)
\(M=\mathbf{4 0 0} \mathrm{g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)\)
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