$A$ 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

If the seventh harmonic of a closed organ pipe is in unison with the fourth harmonic of an open organ pipe, then the ratio of the length of the closed pipe to that of the open pipe is:

In a closed organ pipe of length $105 \,cm$,standing waves are set up corresponding to the third overtone. What is the distance from the closed end where a pressure node is formed? (in $cm$)

In a resonance tube,the first and second resonances are obtained at depths $22.7 \,cm$ and $70.2 \,cm$ respectively. The third resonance will be obtained at a depth of: (in $\,cm$)

$A$ 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 $f$ 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).