The fundamental frequency of a closed pipe is $300 \ Hz$. What will be the frequency of its second overtone (in $Hz$)?

$A$ closed organ pipe and an open organ pipe of the same length produce $2 \text{ beats/second}$ while vibrating in their fundamental modes. The length of the open organ pipe is halved and that of the closed pipe is doubled. Then the number of beats produced per second while vibrating in the fundamental mode is

Tube $A$ has both ends open while tube $B$ has one end closed,otherwise they are identical. The ratio of the fundamental frequency of tube $A$ to tube $B$ is

$A$ closed organ pipe has a fundamental frequency of $1.5\, kHz$. The number of overtones that can be distinctly heard by a person with this organ pipe will be: (Assume that the highest frequency a person can hear is $20,000\, Hz$)

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 ..... $m$.

$A$ tuning fork of frequency $340 \, Hz$ is vibrated just above a cylindrical tube of length $120 \, cm$. Water is slowly poured into the tube. If the speed of sound is $340 \, m/s$,then the minimum height of water required for resonance is .... $cm$.