Question

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 whose Young's modulus is :

• A. $Y=\frac{Y_1{Y}_2}{2(Y_1+Y_2)}$
• B. $Y=\frac{2Y_1{Y}_2}{3(Y_1+Y_2)}$
• C. $Y=\frac{Y_1{Y}_2}{Y_1+Y_2}$
• D. $Y=\frac{2Y_1{Y}_2}{Y_1+Y_2}$

Answer 1

The correct option is D $Y=\frac{2Y_1{Y}_2}{Y_1+Y_2}$

For two wires joined in series $\Delta l=l_1+l_2$

Now, $Y=\frac{\frac{F}{A}}{\frac{\Delta l}{l}}\Rightarrow \Delta l=\frac{Fl}{AY}$


Equivalent length of rod, after joining is $=2l$
As, their lengths are same and force is also same in series.
$\Delta l=\Delta l_1+\Delta l_2$

$\frac{F{l}_{eq}}{A{Y}_{eq}}=\frac{Fl}{A{Y}_1}+\frac{Fl}{A{Y}_2}$

$\frac{l_{eq}}{Y_{eq}}=\frac{l}{Y_1}+\frac{l}{Y_2}$

$\Rightarrow \frac{2l}{Y}=\frac{l}{Y_1}+\frac{l}{Y_2}$

Solving the above equation, we get,
$\therefore Y=\frac{2Y_1{Y}_2}{Y_1+Y_2}$

Hence, $\left(B\right)$ is the correct answer.