To increase the frequency of a stretched string from $100 \ Hz$ to $400 \ Hz$,the tension in the string has to be changed by how many times?

An iron load of $2 \ kg$ is suspended in air from the free end of a sonometer wire of length $1 \ m$. $A$ tuning fork of frequency $256 \ Hz$ is in resonance with $\frac{1}{\sqrt{7}}$ times the length of the sonometer wire. If the load is immersed in water,the length of the wire in metres that will be in resonance with the same tuning fork is (Specific gravity of iron $= 8$).

$A$ wire of density $8 \times 10^3\,kg/m^3$ is stretched between two clamps $0.5\,m$ apart. The extension developed in the wire is $3.2 \times 10^{-4}\,m$. If Young's modulus $Y = 8 \times 10^{10}\,N/m^2$,the fundamental frequency of vibration in the wire will be $......\,Hz$.

$A$ steel wire of length $1 \,m$, mass $0.1 \,kg$, and uniform area of cross-section $10^{-6} \,m^2$ is rigidly fixed at both ends without any tension. Its temperature is lowered by $20^{\circ} C$ and transverse waves are set up by plucking the wire at the middle. The frequency of the fundamental mode is ($Y = 200 \,GPa$, $\alpha = 1.21 \times 10^{-5} {}^{\circ} C^{-1}$). (in $\,Hz$)

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:

The fundamental frequency of a sonometer wire is $n$. If the length and diameter of the wire are doubled while keeping the tension constant,what is the new fundamental frequency?