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 $330 \mathrm{~m} / \mathrm{s}$ )
JEE MAIN 2024, Diffcult
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$ \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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