Solution:

Let $Y Z$ be a mild steel wire with a cross sectional area of $A=0.60 \times 10^{-2} cm^{2}$ and a length of 2m. As indicated in the image, a mass of $m=100g=0.1kg$ is hanged from the midpoint 0. Let $x$ be the depression at the midpoint of the curve, i.e. $OD$.

From the figure, we see,

$Z O=Y O=1=1 \mathrm{~m}$

$\mathrm{M}=0.1 \mathrm{KG}$

$Z D=Y D=\left(I^{2}+x^{2}\right)^{1 / 2}$

Increase in length, $\Delta I=Y D+D Z-Z Y$

$=2 Y D-Y Z$

$(A s D Z=Y D)$

$=2\left(k^{2}+x^{2}\right)^{1 / 2}-2 \mid$

$\Delta \mid=2 l\left(x^{2} /\left.2\right|^{2}\right)=x^{2} / 1$

Therefore, Iongitudinal strain $=\Delta \mathrm{l} / 2 \mathrm{l}=\mathrm{x}^{2} /\left.2\right|^{2} \ldots \ldots$ (i)

If $T$ is the tension in the wires, then in equilibrium $2 T \cos \theta=2 \mathrm{mg}$

, $\mathrm{T}=\mathrm{mg} / 2 \cos \theta$

Or,

$=\left[m g\left(\left.\right|^{2}+x^{2}\right)^{1 / 2}\right] / 2 x=m g l / 2 x$

Therefore, Stress $=\mathrm{T} / A=m g \mid / 2 A x \quad \ldots \ldots \ldots \ldots$. (ii)

$Y=\frac{s t r e s s}{s t r a i n}=\frac{m g l}{2 A x} \times \frac{2 l^{2}}{x^{2}}$

$=\frac{m g l^{3}}{2 A x^{3}}$

$\mathrm{x}=l\left[\frac{\mathrm{mg}}{\mathrm{YA}}\right]^{\frac{1}{3}}=1\left[\frac{0.1 \times 10}{20 \times 10^{11} \times 0.6 \times 10^{-6}}\right]^{\frac{1}{3}}$

$=9.41 \times 10^{-3} \mathrm{~m}$