$A$ 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 remains the same. The frequency of the fork is ............. $Hz$.

$A$ sonometer wire resonates with a given tuning fork forming standing waves with five antinodes between the two bridges when a mass of $9 \,kg$ is suspended from the wire. When this mass is replaced by a mass $M$, the wire resonates with the same tuning fork forming three antinodes for the same positions of the bridges. The value of $M$ is

Two wires made up of the same material are of equal lengths but their radii are in the ratio $1 : 2$. On stretching each of these two strings by the same tension,the ratio between the fundamental frequencies 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

The fundamental frequency of a sonometer wire increases by $9 \ Hz$,if its tension is increased by $69 \%$,keeping the length constant. The frequency of the wire is (in $Hz$)

$A$ segment of wire vibrates with a fundamental frequency of $450 \,Hz$ under a tension of $9 \,kg-wt$. The tension at which the fundamental frequency of the same wire becomes $900 \,Hz$ is: