Two closed pipes produce $10$ beats per second when emitting their fundamental notes. If their lengths are in the ratio of $25 : 26$,then their fundamental frequencies in $Hz$ are:

If the length of an open organ pipe is $33.3 \,cm$, then the frequency of the fifth overtone is [Neglect end correction, velocity of sound $= 333 \,m/s$]. (in $\,Hz$)

The fundamental note produced by a closed organ pipe is of frequency $f$. The fundamental note produced by an open organ pipe of the same length will be of frequency .... $(f)$

$A$ source of frequency $340 \,Hz$ is kept above a vertical cylindrical tube closed at the lower end. The length of the tube is $120 \,cm$. Water is slowly poured in just enough to produce resonance. Then, the minimum height (velocity of sound $= 340 \,m/s$) of the water level in the tube for that resonance is (in $\,m$)

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 $\,cm$)

An open pipe is in resonance in its $2^{nd}$ harmonic with a tuning fork of frequency ${f_1}$. Now,it is closed at one end. If the frequency of the tuning fork is increased slowly from ${f_1}$,then again a resonance is obtained with a frequency ${f_2}$. If in this case the pipe vibrates in its $n^{th}$ harmonic,then: