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 $\,N$)

$A$ string is stretched between two fixed points separated by $75\,cm$. It is observed to have resonant frequencies of $420\,Hz$ and $315\,Hz$. There are no other resonant frequencies between these two. Then,the lowest resonant frequency for this string is ..... $Hz$.

$A$ string of length $2 \ m$ is fixed at both ends. If this string vibrates in its fourth normal mode with a frequency of $500 \ Hz$,then the waves would travel on it with a velocity of ..... $m/s$.

$A$ steel wire of length $1.25 \ m$ is stretched between two rigid supports. The tension in the wire produces an elastic strain of $0.14 \%$. The fundamental frequency of the wire is (Density and Young's modulus of steel are $7.7 \times 10^3 \ kg \ m^{-3}$ and $2.2 \times 10^{11} \ N \ m^{-2}$ respectively). (in $Hz$)

$A$ 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:

$A$ sonometer wire of length $1.5 \ m$ is made of steel. The tension in it produces an elastic strain of $1 \%$. What is the fundamental frequency of steel if density and elasticity of steel are $7.7 \times 10^3 \ kg/m^3$ and $2.2 \times 10^{11} \ N/m^2$ respectively (in $Hz$)?