When an open pipe is closed from one end, the second overtone of the closed pipe is higher in frequency by $100 \text{ Hz}$ than the first overtone of the open pipe. The fundamental frequency of the open pipe will be (Neglect end correction). (in $\text{ Hz}$)

$A$ tuning fork is vibrating at $250\, {Hz}$. The length of the shortest closed organ pipe that will resonate with the tuning fork will be ..... ${cm}$.
(Take speed of sound in air as $340\, {ms}^{-1}$)

$A$ cylindrical tube $(L = 120 \, cm)$ is resonant with a tuning fork of frequency $330 \, Hz$. If it is being filled with water,then to get resonance again,the minimum length of the water column is ...... $cm$ $(v_{air} = 330 \, m/s)$.

An open pipe of length $l$ is vibrating in the $3^{rd}$ overtone with a maximum amplitude $A$. The amplitude at a distance of $\frac{l}{16}$ from any open end is

The waves set up in a closed pipe are

$A$ closed organ pipe of length $L_c$ and an open organ pipe of length $L_o$ contain different gases of densities $\rho_1$ and $\rho_2$ respectively. The compressibility of the gases is the same in both the pipes. The gases are vibrating in their first overtone with the same frequency. What is the length of the open organ pipe?