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(C) For wires joined in series,the total extension $\Delta l$ is the sum of individual extensions: $\Delta l = \Delta l_{1} + \Delta l_{2}$.From the d

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$Y = \frac{2 Y_{1} Y_{2}}{Y_{1} + Y_{2}}$

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(C) For wires joined in series,the total extension $\Delta l$ is the sum of individual extensions: $\Delta l = \Delta l_{1} + \Delta l_{2}$.From the definition of Young's modulus,$Y = \frac{F/A}{\Delta l/l}$,we have $\Delta l = \frac{Fl}{AY}$.Since the wires are joined in series,the force $F$ and cross-sectional area $A$ are the same for both wires.For the equivalent wire of length $2l$ and Young's modulus $Y$,the total extension is $\Delta l = \frac{F(2l)}{AY}$.Substituting the expressions for $\Delta l$,$\Delta l_{1}$,and $\Delta l_{2}$ into the sum equation:$\frac{F(2l)}{AY} = \frac{Fl}{AY_{1}} + \frac{Fl}{AY_{2}}$.Dividing both sides by $Fl/A$,we get:$\frac{2}{Y} = \frac{1}{Y_{1}} + \frac{1}{Y_{2}}$.$\frac{2}{Y} = \frac{Y_{1} + Y_{2}}{Y_{1} Y_{2}}$.Therefore,$Y = \frac{2 Y_{1} Y_{2}}{Y_{1} + Y_{2}}$.

Two wires of the same length and radius are joined end-to-end and loaded. The Young's moduli of the materials of the two wires are $Y_{1}$ and $Y_{2}$. If the combination behaves as a single wire,then its equivalent Young's modulus is:

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