$A$ 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 $; m$)

$A$ sonometer wire resonates with $4$ antinodes between the two bridges for a given tuning fork when a $1 \text{ kg}$ mass is suspended from the wire. Using the same fork, when mass $M$ is suspended, the wire resonates producing $2$ antinodes between the two bridges. (The distance between the bridges remains the same). The value of $M$ is: (in $\text{ kg}$)

$A$ piano wire with a diameter of $0.90 \ mm$ is replaced by another wire of diameter $0.93 \ mm$ of the same material. If the tension of the wire is kept the same,then the percentage change in the frequency of the fundamental tone is

If the length of a stretched string is shortened by $x \%$ and the tension is increased by $44 \%$,then the ratio of the initial and final fundamental frequencies is $1:2$. The value of '$x$' is:

If the frequency of vibrations of a string is to be increased by a factor of two,then the tension in the string must be made:

$A$ sonometer wire has a length of $114 \ cm$, between two fixed ends. Where should two bridges be placed so as to divide the wire into three segments (in $cm$) whose fundamental frequencies are in the ratio $1:3:4$?