$A$ wire stretched between two rigid supports vibrates in its fundamental mode with a frequency of $25 \ Hz$. The mass of the wire is $2 \ g$ and its linear mass density is $4 \times 10^{-3} \ kg/m$. What is the tension in the string (in $N$)?

$A$ 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 $1:3:4$?

Two wires $A$ and $B$ of lengths in the ratio $1: 2$ and masses in the ratio $2: 1$ are stretched by the same tension. The ratio of the fundamental frequencies of wires $A$ and $B$ is

$A$ stone is hung in air from a wire which is stretched over a sonometer. The bridges of the sonometer are $L \, cm$ apart when the wire is in unison with a tuning fork of frequency $N$. When the stone is completely immersed in water,the length between the bridges is $l \, cm$ for re-establishing unison. The specific gravity of the material of the stone is:

In order to double the frequency of the fundamental note emitted by a stretched string, the length is reduced to $\frac{3}{4}$ of the original length and the tension is changed. The factor by which the tension is to be changed is

$A$ metallic wire of length $L$ is fixed between two rigid supports. If the wire is cooled through a temperature difference $\Delta T$ ($Y =$ Young's modulus,$\rho =$ density,$\alpha =$ coefficient of linear expansion),then the frequency of transverse vibration is proportional to: