$A$ source of frequency $340 \,Hz$ is kept above a vertical cylindrical tube closed at the lower end. The length of the tube is $120 \,cm$. Water is slowly poured in just enough to produce resonance. Then, the minimum height (velocity of sound $= 340 \,m/s$) of the water level in the tube for that resonance is (in $\,m$)

An open and a closed organ pipe have the same length. The ratio of the $p^{th}$ mode of frequency of vibration of the two pipes is

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 $330 \,m/s$)

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 $(l_1 \ll l)$

The fundamental frequency of a closed pipe is $220 \ Hz$. If $\frac{1}{4}$ of the pipe is filled with water,the frequency of the first overtone of the pipe now is ..... $Hz$.

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: