A wire density $9\times {10}^3 kg/m^3$   is stretched between two clamps 1 m apart and is subjected to an extension of   $4.9\times {10}^{-4}m.$ The lowest frequency of transverse vibration in the wire is $(Y=9\times {10}^{10}N/m^2) \frac{x}{5}Hz,$  the value of  $x$ is

For wire if
$M=mass,\rho =density,A=Areaofcross\sec tion$
$V=volume,l=length,\Delta l=changeinlength$
Then mass per unit length  $m=\frac{M}{l}=\frac{Al\rho }{l}=A\rho$
And Young’s modulus of elasticity  $Y=\frac{T/A}{\Delta l/l}$
$\Rightarrow T=\frac{Y\Delta lA}{l}.$
Hence lowest frequency of vibration
$n=\frac{1}{2l}\sqrt{\frac{T}{m}}=\frac{1}{2l}\sqrt{\frac{Y\left(\frac{\Delta l}{l}\right)A}{A\rho }}=\frac{1}{2l}\sqrt{\frac{Y\Delta l}{l\rho }}$
$\Rightarrow n=\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{1\times 9\times {10}^3}}=35Hz$