An organ pipe closed at one end produces a fundamental note of frequency '$\nu$'. The pipe is cut into two pipes of equal length. The fundamental frequencies produced in the two pipes are

$A$ pipe open at both ends and a pipe closed at one end have the same length. The ratio of the frequencies of their $P^{\text{th}}$ overtone is:

Two closed pipes have the same fundamental frequency. One is filled with oxygen and the other with hydrogen at the same temperature. The ratio of their lengths respectively is

In the resonance tube experiment,the first resonant length is $l_1$ and the second resonant length is $l_2$. Then,the third resonant length will be:

The first overtone of the vibrating air column of a closed pipe has the same frequency as the second overtone of an open pipe $1.5 \ m$ long. The length of the closed pipe is (in $m$)

In a closed organ pipe,the frequency of the fundamental note is $30 \,Hz$. $A$ certain amount of water is now poured into the organ pipe so that the fundamental frequency increases to $110 \,Hz$. If the organ pipe has a cross-sectional area of $2 \,cm^2$,the amount of water poured into the organ tube is . . . . . . $g$. (Take the speed of sound in air as $330 \,m/s$)