There are two wires of same material and same length while the diameter of second wire is 2 times the diameter of first wire, then ratio of extension produced in the wires by applying same load will be

The correct option is 4:1.

Step-by-step Explanation:

According to Hooke's Law and the definition of Young's modulus ($Y$), the relationship between the applied force ($F$), original length ($L$), cross-sectional area ($A$), and the extension ($\Delta L$) is given by the formula:
$Y=\frac{F/A}{\Delta L/L}$

Rearranging the equation to solve for the extension ($\Delta L$):
$\Delta L=\frac{F\cdot L}{A\cdot Y}$

The cross-sectional area of a wire in terms of its diameter ($d$) is:
$A=\frac{\pi \cdot d^2}{4}$

Substituting the area formula back into the extension formula gives:
$\Delta L=\frac{4\cdot F\cdot L}{\pi \cdot d^2\cdot Y}$

Since both wires are made of the same material (same Young's modulus $Y$), have the same length ($L$), and are subjected to the same load ($F$), the extension is inversely proportional to the square of the diameter:
$\Delta L\propto \frac{1}{d^2}$

Thus, the ratio of the extension produced in the first wire to the second wire is:
$\frac{\Delta L_1}{\Delta L_2}={\left(\frac{d_2}{d_1}\right)}^2$

Given that the diameter of the second wire is 2 times the diameter of the first wire ($d_2=2d_1$):
$\frac{\Delta L_1}{\Delta L_2}={\left(\frac{2d_1}{d_1}\right)}^2=2^2=4$

Therefore, the ratio of extension produced in the wires is 4:1.