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(B) The two wires are connected in parallel to the same load. Let the length of each wire be $L$ and cross-sectional area be $A$.When a total force $F

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$\frac{Y_1 + Y_2}{2}$

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(B) The two wires are connected in parallel to the same load. Let the length of each wire be $L$ and cross-sectional area be $A$.When a total force $F$ is applied,the extension $\Delta L$ in both wires is the same.The force in each wire is $F_1 = \frac{Y_1 A \Delta L}{L}$ and $F_2 = \frac{Y_2 A \Delta L}{L}$.The total force $F = F_1 + F_2 = \frac{(Y_1 + Y_2) A \Delta L}{L}$.If $Y_{eq}$ is the equivalent Young's modulus,then $F = \frac{Y_{eq} (2A) \Delta L}{L}$ (since the total area is $2A$).Equating the two expressions for $F$: $\frac{Y_{eq} (2A) \Delta L}{L} = \frac{(Y_1 + Y_2) A \Delta L}{L}$.Solving for $Y_{eq}$,we get $Y_{eq} = \frac{Y_1 + Y_2}{2}$.

Two wires of equal length and cross-section are suspended as shown. Their Young's moduli are $Y_1$ and $Y_2$ respectively. The equivalent Young's modulus will be

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