Two persons pull a wire towards themselves. Each person exerts a force of $200 \mathrm{~N}$ on the wire. Young's modulus of the material of wire is $1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$. Original length of the wire is $2 \mathrm{~m}$ and the area of cross section is $2 \mathrm{~cm}^2$. The wire will extend in length by _________ $\mu \mathrm{m}$.

<p>To determine the extension in the length of the wire, we can use the formula derived from Young's modulus:</p> <p>$\text{Young's Modulus (Y)} = \frac{\text{Stress}}{\text{Strain}}$</p> <p>Where:</p> <p>Stress ($\sigma$) is given by:</p> <p>$\sigma = \frac{F}{A}$</p> <p>and Strain ($\epsilon$) is:</p> <p>$\epsilon = \frac{\Delta L}{L}$</p> <p>Here,</p> <ul> <li>$F$ is the force exerted,</li> <li>$A$ is the cross-sectional area,</li> <li>$\Delta L$ is the change in length,</li> <li>$L$ is the original length.</li> </ul> <p>We are given:</p> <ul> <li>$F = 200 \mathrm{~N}$ (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),</li> <li>$A = 2 \mathrm{~cm}^2 = 2 \times 10^{-4} \mathrm{~m}^2$,</li> <li>$L = 2 \mathrm{~m}$,</li> <li>$Y = 1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}$.</li> </ul> <p>First, calculate the stress ($\sigma$):</p> <p>$$ \sigma = \frac{200 \mathrm{~N}}{2 \times 10^{-4} \mathrm{~m}^2} = 10^6 \mathrm{~N} \mathrm{~m}^{-2} $$</p> <p>Using Young’s modulus formula:</p> <p>$$ Y = \frac{\text{Stress}}{\text{Strain}} \implies \text{Strain} = \frac{\text{Stress}}{Y} $$</p> <p>Thus, the strain ($\epsilon$) is:</p> <p>$$ \epsilon = \frac{10^6 \mathrm{~N} \mathrm{~m}^{-2}}{1 \times 10^{11} \mathrm{~N} \mathrm{~m}^{-2}} = 10^{-5} $$</p> <p>Now, relate the strain to the change in length:</p> <p>$$ \epsilon = \frac{\Delta L}{L} \implies \Delta L = \epsilon \times L = 10^{-5} \times 2 \mathrm{~m} = 2 \times 10^{-5} \mathrm{~m} $$</p> <p>Convert this change in length to micrometers:</p> <p>$1 \mathrm{~m} = 10^6 \mu \mathrm{m}$</p> <p>$$ 2 \times 10^{-5} \mathrm{~m} = 2 \times 10^{-5} \times 10^6 \mu \mathrm{m} = 20 \mu \mathrm{m} $$</p> <p>Therefore, the wire will extend in length by $20 \mu \mathrm{m}$.</p>