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 $330 \mathrm{~m} / \mathrm{s}$)
Answer (integer)
400
Solution
<p>$$\begin{aligned}
& \frac{V}{4 \ell_1}=30 \Rightarrow \ell_1=\frac{11}{4} m \\
& \frac{V}{4 \ell_2}=110 \Rightarrow \ell_2=\frac{3}{4} m \\
& \Delta \ell=2 m,
\end{aligned}$$</p>
<p>Change in volume $=A \Delta \ell=400 \mathrm{~cm}^3$</p>
<p>$M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)$</p>
About this question
Subject: Physics · Chapter: Waves · Topic: Wave Motion and Types
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