Waves And Sound Question 3
Question 3 - 2024 (30 Jan Shift 1)
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 $330 \mathrm{~m} / \mathrm{s}$ )
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Answer: (400)
Solution:
$\frac{V}{4 \ell_{1}}=30 \Rightarrow \ell_{1}=\frac{11}{4} m$
$\frac{V}{4 \ell_{2}}=110 \Rightarrow \ell_{2}=\frac{3}{4} m$
$\Delta \ell=2 m$
Change in volume $=A \Delta \ell=400 \mathrm{~cm}^{3}$
$\mathbf{M}=\mathbf{4 0 0} \mathrm{g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^{3}\right)$
