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, $2 T=m g$
i.e., $T=\frac{m g}{2}$
Hence, stress on each wire $=\frac{T}{A}=\frac{m g}{2 A}$
$\therefore Y_{1}=\frac{\text { stress }}{\text { strain }}=\frac{\frac{m g}{2 A}}{\frac{\Delta l}{l}}=\frac{m g l}{2 A \Delta l}$ ......(i)
and $Y_{2}=\frac{\text { stress }}{\text { strain }}=\frac{\frac{m g}{2 A}}{\frac{\Delta l}{l}}=\frac{m g l}{2 A \Delta l}$......(ii)
If $Y$ be the equivalent Young's modulus of the combination, then,
$ Y=\frac{m g l}{A \Delta l}.......(iii)$
From Eqs. (i), (ii) and (iii), we get
$Y=\frac{Y_{1}+Y_{2}}{2}$