$A$ string in a musical instrument is $50 \,cm$ long and its fundamental frequency is $800 \,Hz$. Keeping the tension applied to the string same, the change in the length to produce a sound note of fundamental frequency $1000 \,Hz$ will be: (in $\,cm$)

$A$ 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 $\,kg$)

$L$ length and $M$ mass uniform wire is stretched between two fixed points with tension $F$. $A$ sound of frequency $\mu$ is impressed on it. The maximum vibrational energy exists in the wire when $\mu$ =

$A$ guitar string of length $90\,cm$ vibrates with a fundamental frequency of $120\,Hz.$ The length of the string producing a fundamental frequency of $180\,Hz$ will be $...........cm$.

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:

$A$ sonometer wire stretched by weight '$w$' is in unison with a tuning fork. The corresponding resonating length is '$L_1$'. If the weight is increased by '$3w$',the corresponding resonating length of the sonometer in unison with the tuning fork becomes '$L_2$'. The ratio $\left(\frac{L_1}{L_2}\right)$ is: