$A$ tuning fork vibrating with a sonometer having a wire of length $20 \,cm$ produces $5$ beats per second. The beat frequency does not change if the length of the wire is changed to $21 \,cm$. The frequency of the tuning fork must be ............ $Hz$.

Two wires are fixed in a sonometer. Their tensions are in the ratio $8:1$. The lengths are in the ratio $36:35$. The diameters are in the ratio $4:1$. Densities of the materials are in the ratio $1:2$. If the lower frequency in the setting is $360 \ Hz$,the beat frequency when the two wires are sounded together is: (in $Hz$)

$A$ uniform wire of length $L$,diameter $D$,and density $\rho$ is stretched under a tension $T$. The correct relation between its fundamental frequency $f$,the length $L$,and the diameter $D$ is:

In a sonometer wire,the tension is maintained by suspending a $50.7 \, kg$ mass from the free end of the wire. The suspended mass has a volume of $0.0075 \, m^3$. The fundamental frequency of the wire is $260 \, Hz$. If the suspended mass is completely submerged in water,the fundamental frequency will become .... $Hz$ (take $g = 10 \, ms^{-2}$)

When a vibrating tuning fork is placed on a sound box of a sonometer, $8$ beats per second are heard when the length of the sonometer wire is kept at $101 \,cm$ or $100 \,cm$. Then the frequency of the tuning fork is (Consider that the tension in the wire is kept constant). (in $\,Hz$)

Two similar sonometer wires have fundamental frequencies of $500 \, Hz$. They are under the same tension. By what percentage should the tension be increased in one wire so that the two wires produce $5 \, \text{beats/sec}$ (in $\%$)?