QJEEM23Phb2ce: JEE Main 2023 Physics - Class 11 JEEM23Phb2ce Solution: Young's moduli of the material of wires A and B are in the…

Given the formula for elongation in a material due to a force:

$\Delta L = \frac{FL}{AY}$

where:

• F is the force applied,
• L is the original length,
• A is the cross-sectional area of the material, and
• Y is Young's modulus of the material.

The ratio of the elongations in the two wires A and B is given by:

$\frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2} \times \frac{A_2}{A_1} \times \frac{Y_2}{Y_1}$

Since the same force is applied on both wires (i.e., ( $F_1/F_2$ = 1)), the areas are in the ratio 1:3 (i.e., ( $A_2/A_1$ = 3)), and the Young's moduli are in the ratio 1:4 (i.e., ( $Y_2/Y_1$ = 4)), substituting these values into the equation gives:

$\frac{\Delta L_1}{\Delta L_2} = 1 \times 3 \times 4 = 12$

So, the ratio of the elongations is 12:1, which indicates that wire A will elongate 12 times more than wire B when the same force is applied.