The air columns in two tubes closed at one end vibrating in their fundamental modes produce $2$ beats per second. The number of beats produced per second when the same tubes are vibrated in their fundamental mode with their both ends open are

Two tuning forks $A$ and $B$ are sounded together giving rise to $8$ beats in $2$ s. When fork $A$ is loaded with wax, the beat frequency is reduced to $4$ beats in $2$ s. If the original frequency of tuning fork $B$ is $380$ Hz, then the original frequency of tuning fork $A$ is . . . . . . Hz.

$A$ tuning fork $A$ of frequency $200 \ Hz$ is sounded with fork $B,$ the number of beats per second is $5.$ By putting some wax on $A,$ the number of beats increases to $8.$ The frequency of fork $B$ is .... $Hz$

Ten tuning forks are arranged in increasing order of frequency in such a way that any two nearest tuning forks produce $4 \text{ beats/sec}$. The highest frequency is twice the lowest. The possible highest and lowest frequencies are:

$A$ tuning fork arrangement (pair) produces $4 \, beats/sec$ with one fork of frequency $288 \, cps$. $A$ little wax is placed on the unknown fork and it then produces $2 \, beats/sec$. The frequency of the unknown fork is .... $cps$.

$A$ set of $14$ tuning forks is arranged in a series of increasing frequencies. Each fork produces '$x$' beats per second with the preceding fork and the last fork is an octave of the first fork. If the seventh fork has the frequency of $114 \text{ Hz}$, the value of $x$ is