A wire of density  $9\times {10}^{-3}\hspace{0.17em}\mathrm{kg}\hspace{0.17em}{\mathrm{cm}}^{-3}$  is stretched between two clamps 1 m apart. The resulting strain in the wire is  $\text{4.9}\times {\text{10}}^{-4}$ . The lowest frequency of the transverse vibrations in the wire is ______ Hz. (Round-off to the nearest integer)
(Young’s modulus of wire, $Y=9\times {10}^{10} {\mathrm{Nm}}^{-2}$  )

Given 
Density of wire,  $\sigma =9\times {10}^{-3}kgc{m}^{-3}$
Youngs modulus of wire,  $Y=9\times {10}^{10}N{m}^{-2}$
Strain $=4.9\times {10}^{-4}$
 $$\begin{gathered} Y=\frac{Stress}{Strain}=\frac{T/A}{Strain} \\ \therefore \frac{T}{A}=Y\times Strain=9\times {10}^9\times 4.9\times {10}^{-4} \end{gathered}$$
 
Also, mass of wire,  $m=A/\sigma$
Mass per unit length,  $\mu =\frac{m}{J}=A\sigma$
Fundamental frequency in the string
 $$\begin{gathered} f=\frac{1}{2l}\sqrt{\frac{T}{\mu }}=\frac{1}{2l}\sqrt{\frac{T}{\sigma A}} \\ =\frac{1}{2\times 1}\sqrt{\frac{9\times {10}^9\times 4.9\times {10}^{-4}}{9\times {10}^3}} \\ =\frac{1}{2}\sqrt{49\times {10}^{9-4-3}}=\frac{1}{2}\times 70=35Hz \end{gathered}$$