The fundamental frequency of a pipe is $100 \ Hz$ and the next two frequencies are $300 \ Hz$ and $500 \ Hz$. Then,the pipe is:

In a resonance tube experiment,the first resonance is obtained for $10\, cm$ of air column and the second for $32\, cm$. The end correction for this apparatus is ....$cm$

$A$ 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 open at both ends has a fundamental frequency '$n$' in air. The tube is dipped vertically in water so that one-fourth of it is in water. The fundamental frequency of the air column becomes

$A$ closed organ pipe and an open organ pipe are filled with two different gases having the same bulk modulus but different densities $\rho_1$ and $\rho_2$ respectively. The frequency of the $9^{\text{th}}$ harmonic of the closed pipe is identical to the $4^{\text{th}}$ harmonic of the open pipe. If the length of the closed pipe is $10 \ cm$ and the density ratio of the gases is $\rho_1 : \rho_2 = 1 : 16$,then the length of the open pipe is:

If $l_1$ and $l_2$ are the lengths of the air column for the first and second resonance when a tuning fork of frequency $n$ is sounded on a resonance tube,then the distance of the displacement antinode from the top end of the resonance tube is: