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:

$A$ cylindrical tube open at both ends has a fundamental frequency '$f$' in air. The tube is dipped vertically in water so that half of its length is in water. The new fundamental frequency is

$A$ closed organ pipe and an open organ pipe have their first overtones identical in frequency. Their lengths are in the ratio

In a closed organ pipe, the number of nodes formed in the fifth and ninth harmonics are respectively:

$A$ resonance tube closed at one end is of height $1.5 \ m$. $A$ tuning fork of frequency $340 \ Hz$ is vibrating above the tube. Water is poured in the tube gradually. The minimum height of water column for which resonance is obtained is . (Neglect end correction,speed of sound in air $= 340 \ m/s$) (in $cm$)

For a certain organ pipe,the first three resonance frequencies are in the ratio of $1:3:5$ respectively. If the frequency of the fifth harmonic is $405 \, Hz$ and the speed of sound in air is $324 \, ms^{-1}$,the length of the organ pipe is $.......... m$.