In a closed organ pipe, the frequency of fundamental note is 30 ~Hz$30 \mathrm{~Hz}$. A certain amount of water is now poured in the organ pipe so that the fundamental frequency is increased to 110 ~Hz$110 \mathrm{~Hz}$. If the organ pipe has a cross-sectional area of 2 ~cm²$2 \mathrm{~cm}^2$, the amount of water poured in the organ tube is ______ g. (Take speed of sound in air is 330 ~m/s$330 \mathrm{~m/s}$)

Related Formula

Core Logic

For Oxygen (O₂$O_2$) at standard temperature and pressure (S.T.P.): T = 273 K$T = 273 \text{ K}$ M = 32 g/mol = 32 × 10⁻³ kg/mol$M = 32 \text{ g/mol} = 32 \times 10^{-3} \text{ kg/mol}$ γ = 1.4$\gamma = 1.4$ R = 8.3 J K⁻¹ mol⁻¹$R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}$

Step 1: Calculate Velocity

Approximating to the closest given option yields 310 m/s$310 \text{ m/s}$.

Pattern Recognition

For diatomic gases around room temp or STP, velocities range roughly from 250$250$ to 350 m/s$350 \text{ m/s}$ depending on molar mass (N₂ ≈ 334$N_2 \approx 334$, O₂ ≈ 315$O_2 \approx 315$). Recognize 315$315$ is closest to option (1) due to standard approximations taken in exams.

Chapter Mix

Class 11 Physics: Waves Class 11 Physics: Kinetic Theory of Gases

Related Formula

Core Logic

For a closed organ pipe, the fundamental frequency (1st harmonic) is:

where L₁$L_1$ is the length of the closed pipe.

For an open organ pipe, the first overtone (2nd harmonic) is:

where L₂$L_2$ is the length of the open pipe (L₂ = 60 cm$L_2 = 60\mathrm{\,cm}$).

Step 2: Equating Frequencies

Given f₁ = f₂$f_1 = f_2$:

L₂ = 4L₁$L_2 = 4L_1$

Chapter Mix

Class 11 Physics: Waves