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]

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 $.......... m$.

The length of an open organ pipe is $L$ and its fundamental frequency is $f$. If it is now immersed into water up to half of its length,what will be the new fundamental frequency of the organ pipe?

$Assertion :$ The fundamental frequency of an open organ pipe increases as the temperature is increased.
$Reason :$ As the temperature increases,the velocity of sound increases more rapidly than the length of the pipe.

$A$ 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

$A$ uniform tube of length $60.5\,cm$ is held vertically with its lower end dipped in water. $A$ sound source of frequency $500\,Hz$ sends sound waves into the tube. When the length of tube above water is $16\,cm$ and again when it is $50\,cm,$ the tube resonates with the source of sound. Two lowest frequencies (in $Hz$),to which the tube will resonate when it is taken out of water,are (approximately).