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(B) The force constant $k$ of a wire is given by $k = \frac{YA}{L}$.Since the two wires are in parallel,the total force constant $k_{eq}$ is the sum o

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

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(B) The force constant $k$ of a wire is given by $k = \frac{YA}{L}$.Since the two wires are in parallel,the total force constant $k_{eq}$ is the sum of the individual force constants: $k_{eq} = k_1 + k_2$.Substituting the expressions for $k_1$ and $k_2$: $\frac{Y_{eq} (2A)}{L} = \frac{Y_1 A}{L} + \frac{Y_2 A}{L}$.Here,the equivalent system has a total cross-sectional area of $2A$ and the same length $L$.Simplifying the equation: $Y_{eq} (2A) = (Y_1 + Y_2) A$.Therefore,$Y_{eq} = \frac{Y_1 + Y_2}{2}$.

Two wires of the same length and same cross-sectional area are suspended as shown in the figure. Their Young's moduli are $Y_1$ and $Y_2$. What is their equivalent Young's modulus?

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