In a resonance tube experiment,the first and second resonance are heard when the water level is $24.1 \ cm$ and $74.1 \ cm$ respectively,below the open end of the tube. The inner diameter of the tube is (in $cm$)

$n_{1}$ is the frequency of a pipe closed at one end and $n_{2}$ is the frequency of a pipe open at both ends. If both are joined end to end,find the fundamental frequency of the closed pipe so formed.

For a certain organ pipe,three successive resonance frequencies are observed at $425 \, Hz$,$595 \, Hz$,and $765 \, Hz$ respectively. If the speed of sound in air is $340 \, m/s$,then the length of the pipe is ..... $m$.

The fundamental frequency of a closed organ pipe is equal to the first overtone frequency of an open organ pipe. If the length of the open pipe is $60 \,cm$, the length of the closed pipe will be: (in $\,cm$)

$A$ source of sound placed at the open end of a resonance column sends an acoustic wave of pressure amplitude ${\rho _0}$ inside the tube. If the atmospheric pressure is ${\rho _A},$ then the ratio of maximum and minimum pressure at the closed end of the tube will be

An open organ pipe having fundamental frequency $(n)$ is in unison with a vibrating string. If the tube is dipped in water so that $75 \%$ of the length of the tube is inside the water,then the ratio of the fundamental frequency of the air column of the dipped tube to that of the string will be (Neglect end corrections).