To determine the extension in the length of the wire, we can use the formula derived from Young's modulus:

$$ \text{Young's Modulus (Y)} = \frac{\text{Stress}}{\text{Strain}} $$

Where:

Stress ($$\sigma$$) is given by:

$$ \sigma = \frac{F}{A} $$

and Strain ($$\epsilon$$) is:

$$ \epsilon = \frac{\Delta L}{L} $$

Here,

• $$F$$ is the force exerted,
• $$A$$ is the cross-sectional area,
• $$\Delta L$$ is the change in length,
• $$L$$ is the original length.

We are given:

• $$F = 200 \mathrm{~N}$$ (each person pulls with this force, but the total force in the wire should be considered as the tension experienced by one side, so it still remains 200 N),
• $$A = 2 \mathrm{~cm}^2 = 2 \times 10^{-4} \mathrm{~m}^2$$,
• $$L = 2 \mathrm{~m}$$,
• $$Y = 1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$$.

First, calculate the stress ($$\sigma$$):

$$ \sigma = \frac{200 \mathrm{~N}}{2 \times 10^{-4} \mathrm{~m}^2} = 10^6 \mathrm{~N} \mathrm{~m}^{-2} $$

Using Young’s modulus formula:

$$ Y = \frac{\text{Stress}}{\text{Strain}} \implies \text{Strain} = \frac{\text{Stress}}{Y} $$

Thus, the strain ($$\epsilon$$) is:

$$ \epsilon = \frac{10^6 \mathrm{~N} \mathrm{~m}^{-2}}{1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}} = 10^{-5} $$

Now, relate the strain to the change in length:

$$ \epsilon = \frac{\Delta L}{L} \implies \Delta L = \epsilon \times L = 10^{-5} \times 2 \mathrm{~m} = 2 \times 10^{-5} \mathrm{~m} $$

Convert this change in length to micrometers:

$$ 1 \mathrm{~m} = 10^6 \mu \mathrm{m} $$

$$ 2 \times 10^{-5} \mathrm{~m} = 2 \times 10^{-5} \times 10^6 \mu \mathrm{m} = 20 \mu \mathrm{m} $$

Therefore, the wire will extend in length by $$20 \mu \mathrm{m}$$.