When the tension in a string is increased by $3 \ kg \ wt$,the frequency of the fundamental mode increases in the ratio $2:3$. The initial tension in the string is: (in $kg \ wt$)

$A$ stretched string of $1 \ m$ length and mass $5 \times 10^{-4} \ kg$ is having a tension of $20 \ N$. If it is plucked at $25 \ cm$ from one end,then it will vibrate with a frequency of ... $Hz$.

$A$ wire of length $L$ and mass per unit length $6.0 \times 10^{-3} \; kg/m$ is put under a tension of $540 \; N$. Two consecutive frequencies at which it resonates are $420 \; Hz$ and $490 \; Hz$. Then $L$ in meters is: (in $; m$)

$A$ sonometer wire is vibrating in resonance with a tuning fork. Keeping the tension applied same,the length of the wire is doubled. Under what conditions would the tuning fork still be in resonance with the wire?

$A$ uniform string resonates with a tuning fork at a maximum tension of $32 \,N$. If it is divided into two segments by placing a wedge at a distance one-fourth of the length from one end,then to resonate with the same frequency,the maximum value of tension for the string will be ........... $N$.

$A$ sonometer wire is in unison with a tuning fork of frequency '$n$' when it is stretched by a weight of specific gravity '$d$'. When the weight is completely immersed in water,'$x$' beats are produced per second,then