QJEEM23Phdff9: JEE Main 2023 Physics - Class 11 JEEM23Phdff9 Solution: The elastic potential energy stored in a steel wire of length…

Given, energy per unit volume = $\frac{1}{2} \times \text{stress} \times \text{strain}$

The stress can be given as $\text{stress} = Y \times \text{strain}$ , where Y is the Young's modulus.

The energy stored in the wire can be written as:

$\text{Energy} = \frac{1}{2} \times \text{stress} \times \text{strain} \times \text{volume}$

Substituting the stress formula, we get:

$\text{Energy} = \frac{1}{2} \times Y \times \text{strain}^2 \times A \times L$

We are given that the energy stored is $80 \ \text{J}$ , the original length of the wire is $20 \ \text{m}$ , the elongation is $2 \ \text{cm}$ , and the Young's modulus is $2.0 \times 10^{11} \ \text{Nm}^{-2}$ . We need to find the cross-sectional area (A) of the wire.

$80 = \frac{1}{2} \times 2 \times 10^{11} \times \left(\frac{2 \times 10^{-2}}{20}\right)^2 \times A \times 20$

Now we can solve for A:

$A = \frac{80 \times 20^2}{(2.0 \times 10^{11}) \times (2 \times 10^{-2})^2} = 40 \times 10^{-6} \ \text{m}^2$

To convert the area to $\text{mm}^2$ , we multiply by $10^6$ :

$A = 40 \times 10^{-6} \times 10^6 = 40 \ \text{mm}^2$

So, the cross-sectional area of the wire is $40 \ \text{mm}^2$ .