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(A) Let $r_1$ and $r_2$ be the radii of the two wires,then $\frac{r_1}{r_2} = \frac{2}{1}$ (given).We know that stress is defined as $	ext{Stress} =

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$1/4$

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(A) Let $r_1$ and $r_2$ be the radii of the two wires,then $\frac{r_1}{r_2} = \frac{2}{1}$ (given).We know that stress is defined as $	ext{Stress} = \frac{F}{A}$,where $F$ is the force and $A$ is the cross-sectional area.Since the force $F$ applied is the same for both wires,the stress is inversely proportional to the area: $	ext{Stress} \propto \frac{1}{A}$.The area of a circular cross-section is $A = \pi r^2$.Therefore,the ratio of stress produced in the two wires is:$\frac{	ext{Stress}_1}{	ext{Stress}_2} = \frac{A_2}{A_1} = \frac{\pi r_2^2}{\pi r_1^2} = \left(\frac{r_2}{r_1}ight)^2$.Given $\frac{r_1}{r_2} = \frac{2}{1}$,we have $\frac{r_2}{r_1} = \frac{1}{2}$.Substituting this value,we get $\frac{	ext{Stress}_1}{	ext{Stress}_2} = \left(\frac{1}{2}ight)^2 = \frac{1}{4}$.

Consider two wires of the same material having their ratio of radii to be $2:1$. If these two wires are stretched by an equal force,then the ratio of stress produced in them is

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