If the velocity of sound in air is $350 \ m/s$,then the fundamental frequency of an open organ pipe of length $50 \ cm$ will be ............... $Hz$.

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:

For a certain organ pipe,three successive resonance frequencies are observed at $425 \, Hz, 595 \, Hz,$ and $765 \, Hz$ respectively. Taking the speed of sound in air to be $340 \, m/s$,the fundamental frequency of the pipe (in $Hz$) is:

$A$ 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:

In a resonance tube,two successive positions of resonance are obtained at $15 \, cm$ and $48 \, cm$. If the frequency of the tuning fork is $500 \, Hz$,the velocity of sound is ......... $m/s$.

What should be the length of a closed pipe to produce resonance with a sound wave of wavelength $62 \,cm$,in fundamental mode (in $\,cm$)? [Neglect end correction]