Verified Solution • 2025–26 Session moderate

QJEEM24Phf2c8: JEE Main 2024 Physics - Class 11 JEEM24Phf2c8 Solution: In a closed organ pipe, the frequency of fundamental note is .…

In a closed organ pipe, the frequency of fundamental note is 30 Hz30 \mathrm{~Hz} . A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 Hz110 \mathrm{~Hz} . If the organ pipe has a cross-sectional area of 2 cm22 \mathrm{~cm}^2 , the amount of water poured in the organ tube is __________ g. (Take speed of sound in air is 330 m/s330 \mathrm{~m} / \mathrm{s} )

Step-by-Step Verified Solution

amp;V4ℓ1=30⇒ℓ1=114mamp;V4ℓ2=110⇒ℓ2=34mamp;Δℓ=2m,\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 =AΔℓ=400 cm3=A \Delta \ell=400 \mathrm{~cm}^3

M=400 g;(∵ρ=1 g/cm3)M=400 \mathrm{~g} ;\left(\because \rho=1 \mathrm{~g} / \mathrm{cm}^3\right)

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