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.
$\therefore K=\frac{F}{\Delta L}=\frac{YA}{L}$
therefore equivalent spring constant $(K_{eq})$
$K_{eq}=K_1+K_2$
$\frac{Y_{eq}\times 2A}{L}=\frac{Y_{1}A}{L_1}+\frac{Y_{2}A}{L_2}$
Here $L_1$ and $L_2$ are equal
$\therefore Y_{eq}\frac{2A}{L}=\frac{A}{L}(Y_1+Y_2)$
$\Rightarrow Y_{eq}=\frac{Y_1+Y_2}{2}$