Two wires A and B of the same material have r | vedclass

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(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F L}{A \Delta l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-

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$1:1$

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(A) The formula for Young's modulus $(Y)$ is given by $Y = \frac{F L}{A \Delta l}$,where $F$ is the force,$L$ is the original length,$A$ is the cross-sectional area,and $\Delta l$ is the change in length.Rearranging for force,we get $F = \frac{Y A \Delta l}{L}$.Since the material is the same,$Y$ is constant. For the same change in length,$\Delta l$ is constant.Thus,$F \propto \frac{A}{L}$.Since $A = \pi r^2$,we have $F \propto \frac{r^2}{L}$.Given the ratios: $\frac{r_A}{r_B} = \frac{2}{1}$ and $\frac{L_A}{L_B} = \frac{4}{1}$.Therefore,the ratio of forces is $\frac{F_A}{F_B} = \left( \frac{r_A}{r_B} ight)^2 	imes \left( \frac{L_B}{L_A} ight)$.Substituting the values: $\frac{F_A}{F_B} = \left( \frac{2}{1} ight)^2 	imes \left( \frac{1}{4} ight) = 4 	imes \frac{1}{4} = 1$.So,the ratio is $1:1$.

Two wires $A$ and $B$ of the same material have radii in the ratio $2:1$ and lengths in the ratio $4:1$. The ratio of the normal forces required to produce the same change in the lengths of these two wires is

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