Two wires of same length


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 :

  1. $Y=\frac{2 Y_{1} Y_{2}}{3\left(Y_{1}+Y_{2}\right)}$

  2. $\mathrm{Y}=\frac{2 \mathrm{Y}_{1} \mathrm{Y}_{2}}{\mathrm{Y}_{1}+\mathrm{Y}_{2}}$

  3. $Y=\frac{Y_{1} Y_{2}}{2\left(Y_{1}+Y_{2}\right)}$

  4. $\mathrm{Y}=\frac{\mathrm{Y}_{1} \mathrm{Y}_{2}}{\mathrm{Y}_{1}+\mathrm{Y}_{2}}$

Correct Option: , 2


In series combination $\Delta \mathrm{l}=\ell_{1}+\ell_{2}$

$\mathrm{Y}=\frac{\mathrm{F} / \mathrm{A}}{\Delta \ell / \ell} \Rightarrow \Delta \ell=\frac{\mathrm{F} \ell}{\mathrm{AY}}$

$\Rightarrow \Delta \ell \propto \frac{\ell}{\mathrm{Y}}$

Equivalent length of rod after joing is $=2 \ell$

As, lengths are same and force is also same in series

$\Delta \ell=\Delta \ell_{1}+\Delta \ell_{2}$

$\frac{\ell_{\mathrm{cq}}}{\mathrm{Y}_{\mathrm{cq}}}=\frac{\ell}{\mathrm{Y}_{1}}+\frac{\ell}{\mathrm{Y}_{2}} \Rightarrow \frac{2 \ell}{\mathrm{Y}}=\frac{\ell}{\mathrm{Y}_{1}}+\frac{\ell}{\mathrm{Y}_{2}}$

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

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