$A$ 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 .... $Hz$

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 $, Hz$)?

$A$ source of frequency $\nu$ gives $5$ beats/second when sounded with a source of frequency $200 \;Hz$. The second harmonic of frequency $2\nu$ of the source gives $10$ beats/second when sounded with a source of frequency $420 \;Hz$. The value of $\nu$ is .... $Hz$.

When two tuning forks $A$ and $B$ are sounded together, $4$ beats per second are heard. The frequency of the fork $B$ is $384 \,Hz$. When one of the prongs of the fork $A$ is filed and sounded with $B$, the beat frequency increases. Then the frequency of the fork $A$ is: (in $\,Hz$)

$A$ tuning fork sounded together with a tuning fork of frequency $256 \ Hz$ emits $2$ beats per second. On loading the tuning fork of frequency $256 \ Hz,$ the number of beats heard is $1$ per second. The frequency of the unknown tuning fork is: (in $Hz$)

Two tuning forks $A$ and $B$ are sounded together and produce $6$ beats per second. The frequency of tuning fork $B$ is $384 \,Hz$. When tuning fork $A$ is loaded with wax,the number of beats per second becomes $4$. What is the frequency of tuning fork $A$ in $Hz$?