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(D) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the applied force,$L$ is the original leng

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$4 : 1$

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(D) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F \cdot L}{A \cdot \Delta L}$,where $F$ is the applied force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta L$ is the extension.Since both wires are of the same material,their Young's modulus $(Y)$ is the same. Also,the applied load $(F)$ and original length $(L)$ are the same for both wires.Therefore,$A_1 \cdot \Delta L_1 = A_2 \cdot \Delta L_2$.The area of cross-section is $A = \pi r^2$,where $r$ is the radius (or diameter $d/2$).Thus,$r_1^2 \cdot \Delta L_1 = r_2^2 \cdot \Delta L_2$.Given that the diameter of the second wire is twice that of the first,$d_2 = 2d_1$,which implies $r_2 = 2r_1$.Substituting this into the equation: $r_1^2 \cdot \Delta L_1 = (2r_1)^2 \cdot \Delta L_2$.$r_1^2 \cdot \Delta L_1 = 4r_1^2 \cdot \Delta L_2$.$\frac{\Delta L_1}{\Delta L_2} = \frac{4}{1}$.So,the ratio of extension is $4 : 1$.

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

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