$A$ pipe open at both ends has a fundamental frequency $f$ in air. The pipe is now dipped vertically in a water drum to half of its length. The fundamental frequency of the air column is now equal to

$A$ 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

$A$ cylindrical tube open at both ends has a fundamental frequency $f$ in air. When the tube is dipped vertically in water so that one-third part of the tube is in water,the fundamental frequency of the air column becomes (neglect end correction).

An open organ pipe of length $l$ is sounded together with another open organ pipe of length $(l+l_1)$ in their fundamental modes. Speed of sound in air is $V$. The beat frequency heard will be $(l_1 \ll l)$

$A$ pipe,$30.0 \; cm$ long,is open at both ends. Which harmonic mode of the pipe resonates a $1.1 \; kHz$ source? Will resonance with the same source be observed if one end of the pipe is closed? Take the speed of sound in air as $330 \; m s^{-1}$.

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]