JEE MAIN - Physics (2024 - 30th January Morning Shift - No. 26)
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
400
Explanation
$$\begin{aligned} & \frac{V}{4 \ell_1}=30 \Rightarrow \ell_1=\frac{11}{4} m \\ & \frac{V}{4 \ell_2}=110 \Rightarrow \ell_2=\frac{3}{4} m \\ & \Delta \ell=2 m, \end{aligned}$$
Change in volume $$=A \Delta \ell=400 \mathrm{~cm}^3$$
$$M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)$$
Comments (0)
