A wire of density $9\times {10}^3\mathrm{kg}{\mathrm{cm}}^{-3}$ is stretched between two clamps 1 m apart, The resulting strain in the wire is $4.9\times {10}^{-4}$. The lowest frequency (in Hz) of the transverse vibrations in the wire is (Young’s modulus of wire $\mathrm{Y}=9\times {10}^{10}{\mathrm{Nm}}^{-2}$,), (to the nearest integer)_____ 

Densit of wire, $\sigma =9\times {10}^3\mathrm{kg}{\mathrm{cm}}^{-3}$

Young’s modulus of wire, $\mathrm{Y}=9\times {10}^{10}{\mathrm{Nm}}^{-2}$

Strain $4.9\times {10}^{-4}$

$\mathrm{Y}=\frac{\text{ Stress }}{\text{ Strain }}=\frac{\mathrm{T}/\mathrm{A}}{\text{ Strain }}\therefore \frac{\mathrm{T}}{\mathrm{A}}=\mathrm{Y}\times \mathrm{Strain}=9\times {10}^9\times 4.9\times {10}^{-4}$

Also, mass of wire, $m=A\ell \sigma$
Mass per unit length, $\mu =\frac{m}{\ell }=A\sigma$
Fundamental frequency in the string  

$\begin{array}{l}\mathrm{f}=\frac{1}{2\ell }\sqrt{\frac{\mathrm{T}}{\mu }}=\frac{1}{2\ell }\sqrt{\frac{\mathrm{T}}{\sigma \mathrm{A}}}=\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^{10}\times 4.9\times {10}^{-4}}{9\times {10}^3}}\\ =\frac{1}{2}\sqrt{49\times {10}^{10-5-3}}=\frac{1}{2}\times 70=35\mathrm{Hz}\end{array}$