A steel wire with an initial length of 1 m and a cross-sectional...

Let the initial length of the wire be $2l = 1 \text{ m}$, so $l = 0.5 \text{ m}$.

Let $x$ be the vertical depression at the midpoint.

The new length of each half of the wire is $l' = \sqrt{l^2 + x^2}$.

The change in length of each half is $\Delta l = \sqrt{l^2 + x^2} - l \approx l \left(1 + \dfrac{x^2}{2l^2}\right) - l = \dfrac{x^2}{2l}$.

The longitudinal strain in the wire is $\epsilon = \dfrac{\Delta l}{l} = \dfrac{x^2}{2l^2}$.

The tension $T$ in the wire is given by $T = Y A \epsilon = \dfrac{Y A x^2}{2l^2}$.

For equilibrium at the midpoint, the vertical components of the tension balance the weight of the load:

$2 T \sin \theta = m g$

where $\sin \theta = \dfrac{x}{l'} \approx \dfrac{x}{l}$.

Substituting $T$ and $\sin \theta$:

$2 \left( \dfrac{Y A x^2}{2l^2} \right) \left( \dfrac{x}{l} \right) = m g$

$\dfrac{Y A x^3}{l^3} = m g$

$x^3 = \dfrac{m g l^3}{Y A}$

Substituting the given values $m = 2.16 \text{ kg}$, $g = 10 \text{ m/s}^2$, $l = 0.5 \text{ m}$, $Y = 2.0 \times 10^{11} \text{ N/m}^2$, and $A = 4.00 \times 10^{-6} \text{ m}^2$:

$x^3 = \dfrac{2.16 \times 10 \times (0.5)^3}{2.0 \times 10^{11} \times 4.00 \times 10^{-6}}$

$x^3 = \dfrac{21.6 \times 0.125}{8.0 \times 10^5}$

$x^3 = \dfrac{2.7}{8.0 \times 10^5} = \dfrac{27}{8} \times 10^{-6} \text{ m}^3$

Taking the cube root of both sides:

$x = \dfrac{3}{2} \times 10^{-2} \text{ m} = 1.5 \text{ cm}$