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Two wires of same length and radius are joined end to end and loaded. The Young's modulii of the materials of the two wires are $Y_{1}$ and $Y_{2}$. The combination behaves as a single wire then its Young's modulus is:
$y=\frac{Y_{1} Y_{2}}{Y_{1}+Y_{2}}$
$y=\frac{2 Y_{1} Y_{2}}{3\left(Y_{1}+Y_{2}\right)}$
$Y=\frac{2 Y_{1} Y_{2}}{Y_{1}+Y_{2}}$
${Y}=\frac{{Y}_{1} {Y}_{2}}{2\left({Y}_{1}+{Y}_{2}\right)}$
Solution
In series combination $\Delta l =l_{1}+l_{2}$
$Y =\frac{ F / A }{\Delta l / l} \Rightarrow \Delta l=\frac{ F l}{ AY }$
$\Rightarrow \Delta l \propto \frac{l}{ Y }$
Equivalent length of rod after joining is $=2 l$
As, lengths are same and force is also same in series
$\Delta l=\Delta l_{1}+\Delta l_{2}$
$\frac{l \text { eq }}{ Y _{\text {eq }}}=\frac{l}{ Y _{1}}+\frac{l}{ Y _{2}}$ $\Rightarrow \frac{2 l}{ Y }=\frac{l}{ Y _{1}}+\frac{l}{ Y _{2}}$
$\therefore Y=\frac{2 Y_{1} Y_{2}}{Y_{1}+Y_{2}}$
