$A$ mild steel wire of length $1.0 \; m$ and cross-sectional area $0.50 \times 10^{-2} \; cm^{2}$ is stretched,well within its elastic limit,horizontally between two pillars. $A$ mass of $100 \; g$ is suspended from the mid-point of the wire. Calculate the depression at the midpoint.

(0.0106 M) Length of the steel wire $L = 1.0 \; m$.
Area of cross-section $A = 0.50 \times 10^{-2} \; cm^{2} = 0.50 \times 10^{-6} \; m^{2}$.
Mass $m = 100 \; g = 0.1 \; kg$.
Let $l$ be the depression at the midpoint. The wire forms a triangle with the original horizontal position. The new length of each half of the wire is $\sqrt{(L/2)^2 + l^2}$.
Increase in length $\Delta L = 2 \sqrt{(L/2)^2 + l^2} - L = 2(L/2) [1 + (l/(L/2))^2]^{1/2} - L \approx L(1 + l^2/(L/2)^2) - L = 2l^2/L$.
Strain $= \Delta L / L = 2l^2/L^2$.
From force balance at the midpoint,$mg = 2T \sin \theta$,where $\sin \theta = l / \sqrt{(L/2)^2 + l^2} \approx l / (L/2) = 2l/L$.
So,$mg = 2T (2l/L) \implies T = mgL / 4l$.
Stress $= T/A = mgL / (4lA)$.
Young's Modulus $Y = \text{Stress} / \text{Strain} = (mgL / 4lA) / (2l^2/L^2) = mgL^3 / (8Al^3)$.
$l^3 = mgL^3 / (8AY) \implies l = L \sqrt[3]{mg / (8AY)}$.
Using $Y = 2 \times 10^{11} \; Pa$,$l = 1.0 \times \sqrt[3]{(0.1 \times 9.8) / (8 \times 0.50 \times 10^{-6} \times 2 \times 10^{11})} = \sqrt[3]{0.98 / 800000} \approx 0.0106 \; m$.