The ratio of radii of two wires of same material is $2: 1$. If these wires are stretched by equal force, the ratio of stresses produced in them is $.............$
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We know,
$\text { Stress }=\frac{\text { Force }}{\text { Area }}$
So, Stress $\times$ Area $=$ Force
$S \times A=F$
$\left\{\begin{array}{l}S=\text { Stress } \\ F=\text { Force } \\ A=\text { Area } \\ r=\text { radius }\end{array}\right\}$
$\because \ ($Since$)$ Force applied on the wires is equal we can relate two conditions as
$S_1 A_1=S_2 A_2$
$\frac{S_1}{S_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}$
$\frac{S_1}{S_2}=\frac{r^2}{(2 r)^2}=\frac{r^2}{4 r^2}=\frac{1}{4}$
$\left\{\begin{array}{l}\text { Where } \\ S_1-\text { Stress in } 1^{\text {st }} \text { wire } \\ A_1-\text { Area of } 1^{\text {st }} \text { wire } \\ r_1-\text { Radius of } 1^{\text {st }} \text { wire } \\ S_2-\text { Stress in } 2^{\text {nd }} \text { wire } \\ A_2-\text { Area of } 2^{\text {nd }} \text { wire } \\ r_2-\text { Radius of } 2^{\text {nd }} \text { wire }\end{array}\right\}$
$\text { Stress }=\frac{\text { Force }}{\text { Area }}$
So, Stress $\times$ Area $=$ Force
$S \times A=F$
$\left\{\begin{array}{l}S=\text { Stress } \\ F=\text { Force } \\ A=\text { Area } \\ r=\text { radius }\end{array}\right\}$
$\because \ ($Since$)$ Force applied on the wires is equal we can relate two conditions as
$S_1 A_1=S_2 A_2$
$\frac{S_1}{S_2}=\frac{A_2}{A_1}=\frac{\pi r_2^2}{\pi r_1^2}$
$\frac{S_1}{S_2}=\frac{r^2}{(2 r)^2}=\frac{r^2}{4 r^2}=\frac{1}{4}$
$\left\{\begin{array}{l}\text { Where } \\ S_1-\text { Stress in } 1^{\text {st }} \text { wire } \\ A_1-\text { Area of } 1^{\text {st }} \text { wire } \\ r_1-\text { Radius of } 1^{\text {st }} \text { wire } \\ S_2-\text { Stress in } 2^{\text {nd }} \text { wire } \\ A_2-\text { Area of } 2^{\text {nd }} \text { wire } \\ r_2-\text { Radius of } 2^{\text {nd }} \text { wire }\end{array}\right\}$
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