$A$ tuning fork gives $5$ beats per second with $40 \ cm$ length of a 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 $Hz$)

The fundamental frequency of a wire stretched by $2 \ kgwt$ is $100 \ Hz$. The weight required to produce its octave is (in $kgwt$)

$A$ tuning fork gives $4$ beats with $50 \ cm$ length of a 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 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 $\,Hz$)

$A$ wire of length $L$ and mass per unit length $6.0 \times 10^{-3} \; kg/m$ is put under a tension of $540 \; N$. Two consecutive frequencies at which it resonates are $420 \; Hz$ and $490 \; Hz$. Then $L$ in meters is: (in $; m$)

If we add $3 \ kg$ load to the hanger of a sonometer,the fundamental frequency becomes two times its initial value. The initial load must be (in $kg$)