Step-by-step Derivation

1. Understanding the System and SHM: When the mass m is pulled down by a small distance x from its equilibrium position and released, it oscillates. The restoring force is provided by the tension in the wire. This setup results in Simple Harmonic Motion (SHM). The angular frequency ω of this SHM is given by the formula: $ω = \sqrt{\frac{k}{m}}$ where k is the effective spring constant of the system (in this case, the wire).

2. Relating Angular Frequency and Spring Constant: From the SHM formula, we can express the spring constant k in terms of mass m and angular frequency ω: $ω^2 = \frac{k}{m} \implies k = mω^2$

3. Relating Spring Constant and Young's Modulus: The wire acts like a spring. The Young's modulus Y of the material of the wire is defined as the ratio of stress to strain: $Y = \frac{\text{Stress}}{\text{Strain}} = \frac{F/A}{\Delta L/L}$ where F is the force applied, A is the cross-sectional area, L is the original length, and ΔL is the extension.

   Rearranging this formula to find the force F required to produce an extension ΔL: $F = \left(\frac{YA}{L}\right) \Delta L$ This is in the form of Hooke's Law, F = kx, where the displacement x is the extension ΔL. By comparing these two equations, we can identify the effective spring constant k of the wire: $k = \frac{YA}{L}$

4. Solving for Young's Modulus (Y): We now have two expressions for the spring constant k. By equating them, we can solve for Y: $mω^2 = \frac{YA}{L}$ $Y = \frac{mω^2L}{A}$

5. Substituting the Given Values: We are given:

   • Mass, m = 0.1 kg
   • Length, L = 1 m
   • Cross-sectional area, $A = 4.9 × 10^{-7}$ m²
   • Angular frequency, ω = 140 rad s⁻¹

   Now, substitute these values into the equation for Y: $Y = \frac{(0.1 \text{ kg}) \times (140 \text{ rad s}^{-1})^2 \times (1 \text{ m})}{4.9 \times 10^{-7} \text{ m}^2}$ $Y = \frac{0.1 \times 19600}{4.9 \times 10^{-7}} \text{ Nm}^{-2}$ $Y = \frac{1960}{4.9 \times 10^{-7}} \text{ Nm}^{-2}$ $Y = \frac{19600}{49} \times 10^7 \text{ Nm}^{-2}$

6. Final Calculation: Since $196 = 4 \times 49$, we have 19600 / 49 = 400. $Y = 400 \times 10^7 \text{ Nm}^{-2}$ $Y = 4 \times 10^2 \times 10^7 \text{ Nm}^{-2}$ $Y = 4 \times 10^9 \text{ Nm}^{-2}$

7. Determining the Value of n: The problem states that the Young's modulus is $n × 10^9 Nm⁻²$. Comparing this with our calculated value: $n \times 10^9 = 4 \times 10^9$ Therefore, the value of n is 4.