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

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$1:1$

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(B) The formula for Young's modulus $Y$ is given by $Y = \frac{F L}{A \Delta L}$,where $F$ is the load,$L$ is the length,$A$ is the cross-sectional area,and $\Delta L$ is the extension.Given that the load $F$ and the extension $\Delta L$ are the same for both wires,we have $Y \propto \frac{L}{A}$.For wire $A$: $L_A = 5.0 \, m$,$A_A = 2.5 	imes 10^{-5} \, m^2$.For wire $B$: $L_B = 6.0 \, m$,$A_B = 3.0 	imes 10^{-5} \, m^2$.The ratio of Young's moduli is $\frac{Y_A}{Y_B} = \frac{L_A / A_A}{L_B / A_B} = \frac{L_A}{A_A} 	imes \frac{A_B}{L_B}$.Substituting the values: $\frac{Y_A}{Y_B} = \frac{5.0}{2.5 	imes 10^{-5}} 	imes \frac{3.0 	imes 10^{-5}}{6.0} = \frac{5.0}{2.5} 	imes \frac{3.0}{6.0} = 2 	imes 0.5 = 1$.Thus,the ratio is $1:1$.

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