QJEEM24Ph1f25: JEE Main 2024 Physics - Class 11 JEEM24Ph1f25 Solution: Two persons pull a wire towards themselves. Each person exerts…

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 $20 \mu \mathrm{m}$ .