$A$ tuning fork gives $3$ beats with $50 \ cm$ length of sonometer wire. If the length of the wire is shortened by $1 \ cm$,the number of beats is still the same. The frequency of the fork is (in $Hz$)

When two tuning forks (fork $1$ and fork $2$) are sounded simultaneously,$4$ beats per second are heard. Now,some tape is attached to the prong of fork $2$. When the tuning forks are sounded again,$6$ beats per second are heard. If the frequency of fork $1$ is $200 \, Hz$,then what was the original frequency of fork $2$ (in $, Hz$)?

$A$ 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 .... $Hz$

$41$ tuning forks are arranged in such a way that each produces $5 \text{ beats per second}$ when sounded with its nearest fork. If the frequency of the last fork is double the frequency of the first fork,then the frequencies of the first and last fork are respectively:

Two sitar strings $A$ and $B$ playing the note '$Ga$' are slightly out of tune and produce beats of frequency $6 \; Hz$. The tension in the string $A$ is slightly reduced and the beat frequency is found to reduce to $3 \; Hz$. If the original frequency of $A$ is $324 \; Hz$,what is the frequency (in $Hz$) of $B$?

$41$ tuning forks are arranged in increasing order of frequency. Each tuning fork produces $5 \, \text{beats/sec}$ with the next one. If the frequency of the last tuning fork is double that of the first, what are the frequencies of the first and last tuning forks?