1. Given data

• Density of wire: $\rho = 9\times 10^{-3}\ \text{kg cm}^{-3}$
• Length between clamps: $L = 1\ \text{m}$
• Strain: $\epsilon = 4.9\times 10^{-4}$
• Young’s modulus: $Y = 9\times 10^{10}\ \text{N m}^{-2}$

We need the lowest (fundamental) frequency of transverse vibration.

2. Convert density into SI units

Since $1\ \text{cm}^3 = 10^{-6}\ \text{m}^3,$ we have $1\ \text{kg cm}^{-3} = 10^6\ \text{kg m}^{-3}.$

Therefore, $\rho = 9\times 10^{-3}\times 10^6 = 9\times 10^3\ \text{kg m}^{-3}.$

3. Find the tension using stress = Y × strain

Stress in the wire is $\text{stress} = Y\epsilon.$

So, $\text{stress} = 9\times 10^{10}\times 4.9\times 10^{-4}.$

$\text{stress} = 44.1\times 10^6 = 4.41\times 10^7\ \text{N m}^{-2}.$

Now, $\text{stress} = \frac{T}{A}$ where $T$ is tension and $A$ is cross-sectional area.

Thus, $\frac{T}{A} = 4.41\times 10^7.$

4. Use the wave speed formula

For a stretched wire, wave speed is $v = \sqrt{\frac{T}{\mu}},$ where linear mass density $\mu = \rho A.$

Hence, $v = \sqrt{\frac{T}{\rho A}} = \sqrt{\frac{T/A}{\rho}}.$

Substitute values: $v = \sqrt{\frac{4.41\times 10^7}{9\times 10^3}}.$

$v = \sqrt{4.9\times 10^3} = \sqrt{4900} = 70\ \text{m/s}.$

5. Find the fundamental frequency

For a wire fixed at both ends, the lowest frequency is $f_1 = \frac{v}{2L}.$

Since $L=1\ \text{m}$, $f_1 = \frac{70}{2\times 1} = 35\ \text{Hz}.$

6. Final answer

The lowest frequency is $\boxed{35}$

This matches the stored correct answer.