Length of the steel wire $=1.0 m$

Area of cross-section, $A=0.50 \times 10^{-2} cm ^{2}-0.50 \times 10^{-6} m ^{2}$

A mass $100 g$ is suspended from its midpoint.

$m=100 g =0.1 kg$

Hence, the wire dips, as shown in the given figure.

Original length $= XZ$

Depression $=l$

The length after mass $m$, is attached to the wire $= XO + OZ$

Increase in the length of the wire:

$\Delta l=( XO + OZ )- XZ$

$XO = OZ =\left[(0.5)^{2}+l^{2}\right]^{\frac{1}{2}}$

$\therefore \Delta l=2\left[(0.5)^{2}+(l)^{2}\right]^{\frac{1}{2}}-1.0$

$=2 \times 0.5\left[1+\left(\frac{l}{0.5}\right)^{2}\right]^{\frac{1}{2}}-1.0$

Expanding and neglecting higher terms, we get:

$\Delta l=\frac{l^{2}}{0.5}$

Strain $=\frac{\text { Increase in length }}{\text { Original length }}$

Let $T$ be the tension in the wire.

$\therefore m g=2 T \cos \theta$

Using the figure, it can be written as

$\cos \theta=\frac{1}{\left((0.5)^{2}+l^{2}\right)^{\frac{1}{2}}}$

$=\frac{1}{(0.5)\left(1+\left(\frac{l}{0.5}\right)^{2}\right)^{\frac{1}{2}}}$

Expanding the expression and eliminating the higher terms

$\cos \theta=\frac{1}{(0.5)\left(1+\frac{l^{2}}{2(0.5)^{2}}\right)}$

$\left(1+\frac{l^{2}}{0.5}\right)=1$ for small $l$

$\therefore \cos \theta=\frac{l}{0.5}$

$\therefore T=\frac{m g}{2\left(\frac{l}{0.5}\right)}=\frac{m g \times 0.5}{2 l}=\frac{m g}{4 l}$

Stress $=\frac{\text { Tension }}{\text { Area }}=\frac{m g}{4 l \times A}$

Young's modulus $=\frac{\text { Stress }}{\text { Strain }}$

$Y=\frac{m g \times 0.5}{4 l \times A \times l^{2}}$

$I=\sqrt[3]{\frac{m g \times 0.5}{4 Y A}}$

Young's modulus of steel, $Y=2 \times 10^{11} Pa$

$\therefore l=\sqrt{\frac{0.1 \times 9.8 \times 0.5}{4 \times 2 \times 10^{11} \times 0.50 \times 10^{-6}}}$

$=0.0106 m$

Hence, the depression at the midpoint is $0.0106 m$