$A$ closed organ pipe and an open organ pipe of the same length produce $2$ beats per second when they are set into vibrations together in their fundamental mode. The length of the open pipe is now halved and that of the closed pipe is doubled. The number of beats produced per second will be:

$A$ 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

The fundamental frequency of a closed pipe is equal to the frequency of the second harmonic of an open pipe. The ratio of their lengths is

If the length of a closed organ pipe is $1.5 \, m$ and the velocity of sound is $330 \, m/s$,then the frequency for the second note is ... $Hz$.

$A$ tuning fork of frequency $340\,Hz$ resonates in the fundamental mode with an air column of length $125\,cm$ in a cylindrical tube closed at one end. When water is slowly poured in it,the minimum height of water required for observing resonance once again is . . . . . . $cm$. (Velocity of sound in air is $340\,ms^{-1}$)

If the lengths of the open and closed pipes are in the ratio of $2 : 3$, then the ratio of the frequencies of the third harmonic of the open pipe and the fifth harmonic of the closed pipe is