$A$ tuning fork and a sonometer wire are sounded together and produce $4$ beats per second. When the length of the sonometer wire is $95 \ cm$ or $100 \ cm$,the frequency of the tuning fork is ..... $Hz$.

$A$ wire of length $L$,diameter $d$,and density of material $\rho$ is under tension $T$,having a fundamental frequency of vibration $n_A$. Another wire of length $2L$,tension $2T$,density $2\rho$,and diameter $3d$ has a fundamental frequency of vibration $n_B$. The ratio $n_B : n_A$ is

$A$ 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 $; m$)

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:

Two identical strings $X$ and $Z$ made of the same material have tensions $T_{x}$ and $T_{z}$ in them. If their fundamental frequencies are $450\, Hz$ and $300\, Hz$,respectively,then the ratio $T_{x} / T_{z}$ is:

$A$ 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?