$A$ 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:

$A$ 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 strings $A$ and $B$ of lengths $L_A = 80 \text{ cm}$ and $L_B = x \text{ cm}$ respectively are used separately in a sonometer. The ratio of their densities $(d_A / d_B)$ is $0.81$. The diameter of $B$ is one-half that of $A$. If the strings have the same tension and fundamental frequency, the value of $x$ is:

When tension $T$ is applied to a sonometer wire of length $l$, it vibrates with the fundamental frequency $n$. Keeping the setup same, when the tension is increased by $8 \,N$, the fundamental frequency becomes three times the earlier. The initial tension applied to the wire was: (in $\,N$)

The fundamental frequency of a wire stretched by $2 \ kgwt$ is $100 \ Hz$. The weight required to produce its octave is (in $kgwt$)

$A$ uniform string is vibrating with a fundamental frequency '$n$'. If the radius and length of the string are both doubled while keeping the tension constant,then the new frequency of vibration is: