Young's moduli of the material of wires A and B are in the ratio of $1: 4$, while its area of cross sections are in the ratio of $1: 3$. If the same amount of load is applied to both the wires, the amount of elongation produced in the wires $\mathrm{A}$ and $\mathrm{B}$ will be in the ratio of

[Assume length of wires A and B are same]

<p>Given the formula for elongation in a material due to a force:</p> <p>$\Delta L = \frac{FL}{AY}$</p> <p>where:</p> <ul> <li>F is the force applied,</li> <li>L is the original length,</li> <li>A is the cross-sectional area of the material, and</li> <li>Y is Young&#39;s modulus of the material.</li> </ul> <p>The ratio of the elongations in the two wires A and B is given by:</p> <p>$$ \frac{\Delta L_1}{\Delta L_2} = \frac{F_1}{F_2} \times \frac{A_2}{A_1} \times \frac{Y_2}{Y_1} $$</p> <p>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&#39;s moduli are in the ratio 1:4 (i.e., ($Y_2/Y_1$ = 4)), substituting these values into the equation gives:</p> <p>$\frac{\Delta L_1}{\Delta L_2} = 1 \times 3 \times 4 = 12$</p> <p>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.</p>