Question

Two wires of equal length and cross-section area are suspended as shown in figure. Their Young's modulus are $Y_{1}$ and $Y_{2}$ , respectively. The equivalent Young's modulus will be

• A. $Y_{1}+Y_{2}$
• B. $\frac{Y_{1} + Y_{2}}{2}$
• C. $\frac{Y_{1} Y_{2}}{Y_{1} + Y_{2}}$
• D. $\sqrt{Y_{1} Y_{2}}$

Answer 1

Answer: (B)

Given that the wires have equal length and cross section
We know that
$F=K\Delta x$
Where $K$ is spring constant
Let $K_{1}$ and $K_{2}$ be the spring constants of the two wires.
$\therefore K=\frac{F}{\Delta L}=\frac{Y A}{L}$
therefore equivalent spring constant $\left(\right.K_{eq}\left.\right)$
$K_{eq}=K_{1}+K_{2}$
$\frac{Y_{e q} \times 2 A}{L}=\frac{Y_{1} A}{L_{1}}+\frac{Y_{2}A}{L_{2}}$
Here $L_{1}$ and $L_{2}$ are equal
$\therefore Y_{e q}\frac{2 A}{L}=\frac{A}{L}\left(\right.Y_{1}+Y_{2}\left.\right)$
$\Rightarrow Y_{e q}=\frac{Y_{1} + Y_{2}}{2}$