If the end correction of an open pipe is $0.8 \,cm$, then the inner radius of that pipe is

In a resonance tube open at one end, the end correction is $1.1$ cm. If the shortest length of resonating air column with a tuning fork is $18$ cm, the next resonating length will be (in $cm$)

The fundamental frequency of a closed pipe is $400 \,Hz$. If $1/3$ of the pipe is filled with water,the frequency of the $2^{\text{nd}}$ harmonic of the pipe will be (Neglect end correction). (in $\,Hz$)

In a resonance tube of length $0.8 \,m$, an air column vibrates with a source of frequency $375 \,Hz$ for a certain height of water from the bottom of the tube. What is the water level corresponding to the fundamental frequency (in $\,m$)? (Neglect end correction, speed of sound in air = $330 \,m/s$)

The frequency of a vibrating air column in a pipe,open at both ends,is $f_1$. When $\frac{3}{4}$ of its length is immersed in water,the frequency of the vibrating air column is $f_2$. The value of $\frac{f_1}{f_2}$ is

$A$ resonance tube is old and has a jagged end. It is still used in the laboratory to determine the velocity of sound in air. $A$ tuning fork of frequency $512\,Hz$ produces the first resonance when the tube is filled with water to a mark $11\,cm$ below a reference mark,near the open end of the tube. The experiment is repeated with another fork of frequency $256\,Hz$,which produces the first resonance when water reaches a mark $27\,cm$ below the reference mark. The velocity of sound in air,obtained in the experiment,is close to .... $ms^{-1}$