$A$ $1 \, cm$ long string vibrates with a fundamental frequency of $256 \, Hz$. If the length is reduced to $\frac{1}{4} \, cm$ keeping the tension unaltered,the new fundamental frequency will be: (in $, Hz$)

$A$ sonometer wire resonates with $4$ antinodes between two bridges for a given tuning fork, when $1 \,kg$ 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 two bridges remains the same). The value of $M$ is: (in $\,kg$)

$A$ 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 $......\,Hz$.

The correct graph between the frequency $n$ and the square root of density $(\rho)$ of a wire,keeping its length,radius,and tension constant,is:

$A$ tuning fork resonates with a sonometer wire of length $1 \ m$ stretched with a tension of $6 \ N$. When the tension in the wire is changed to $54 \ N$,the same tuning fork produces $12$ beats per second with it. The frequency of the tuning fork is $Hz$.

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: