Question

Two wires of equal length and equal cross sectional areas are suspended as shown in the figure. Their Young's modulii are $Y_1$ and $Y_2$, respectively. The equivalent Young's modulus is

• 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, two wires have same length and equal cross-sectional area.

i.e., $l_1=l_2=l$ and $A_1=A_2=A$
Suppose, $m$ is the mass of the hanging load.
From figure, $2T=mg$
i.e., $T=\frac{mg}{2}$
Hence, stress on each wire $=\frac{T}{A}=\frac{mg}{2A}$
$\therefore Y_1=\frac{\text{ stress }}{\text{ strain }}=\frac{\frac{mg}{2A}}{\frac{\Delta l}{l}}=\frac{mgl}{2A\Delta l}$ ......(i)
and $Y_2=\frac{\text{ stress }}{\text{ strain }}=\frac{\frac{mg}{2A}}{\frac{\Delta l}{l}}=\frac{mgl}{2A\Delta l}$......(ii)
If $Y$ be the equivalent Young's modulus of the combination, then,
$Y=\frac{mgl}{A\Delta l}.......(iii)$
From Eqs. (i), (ii) and (iii), we get
$Y=\frac{Y_1+Y_2}{2}$