Two closed pipes produce $10$ beats per second when emitting their fundamental notes. If their lengths are in the ratio of $25 : 26$,then their fundamental frequencies in $Hz$ are:

$A$ closed organ pipe has length $L$. The air in it is vibrating in the third overtone with a maximum amplitude $a$. The amplitude at a distance $\frac{L}{7}$ from the closed end of the pipe is:

$A$ closed pipe is suddenly opened and changed to an open pipe of the same length. The fundamental frequency of the resulting open pipe is less than that of the $3^{rd}$ harmonic of the earlier closed pipe by $55 \,Hz$. Then, the value of the fundamental frequency of the closed pipe is: (in $\,Hz$)

$A$ 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 pipe of length $33 \, cm$ resonates to a frequency of $1000 \, Hz$. The mode of vibration is: (velocity of sound $= 330 \, m/s$)

$A$ closed organ pipe (closed at one end) is excited to support the third overtone. It is found that air in the pipe has