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(D) The extension $l$ in a wire is given by the formula $l = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the length,$A$ is the cross-sectiona

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$4:1$

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(D) The extension $l$ in a wire is given by the formula $l = \frac{FL}{AY}$,where $F$ is the applied force,$L$ is the length,$A$ is the cross-sectional area,and $Y$ is Young's modulus.Since the material is the same,$Y$ is constant. Given that $F$ and $L$ are also constant,we have $l \propto \frac{1}{A}$.Since $A = \pi r^2$ (where $r$ is the radius),$l \propto \frac{1}{r^2}$.Given that the diameter of the second wire is $2$ times the diameter of the first wire,the radius $r_2 = 2r_1$.Therefore,the ratio of extensions is $\frac{l_1}{l_2} = \left( \frac{r_2}{r_1} \right)^2 = (2)^2 = 4$.Thus,the ratio $l_1 : l_2 = 4 : 1$.

There are two wires of the same material and same length. If the diameter of the second wire is $2$ 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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