The ratio of radii of two wires of same mater | vedclass

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(C) We know that stress is defined as the force applied per unit area:$	ext{Stress} (S) = \frac{F}{A}$Since the area of a circular cross-section is $

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$1: 4$

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(C) We know that stress is defined as the force applied per unit area:$	ext{Stress} (S) = \frac{F}{A}$Since the area of a circular cross-section is $A = \pi r^2$,we have:$S = \frac{F}{\pi r^2}$Given that the force $F$ applied to both wires is equal,the stress is inversely proportional to the square of the radius:$S \propto \frac{1}{r^2}$Therefore,the ratio of stresses $S_1 : S_2$ is:$\frac{S_1}{S_2} = \frac{r_2^2}{r_1^2} = \left( \frac{r_2}{r_1} ight)^2$Given the ratio of radii $r_1 : r_2 = 2 : 1$,we have $\frac{r_2}{r_1} = \frac{1}{2}$.Substituting this value:$\frac{S_1}{S_2} = \left( \frac{1}{2} ight)^2 = \frac{1}{4}$Thus,the ratio of stresses is $1: 4$.

List-$I$	List-$II$
$(A)$ $A$ force that restores an elastic body of unit area to its original state	$(I)$ Bulk modulus
$(B)$ Two equal and opposite forces parallel to opposite faces	$(II)$ Young's modulus
$(C)$ Forces perpendicular everywhere to the surface per unit area same everywhere	$(III)$ Stress
$(D)$ Two equal and opposite force perpendicular to opposite faces	$(IV)$ Shear modulus

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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