A metal wire of unstretched length 4 m and diameter 1 mm suspends a...

Let the original unstretched length of the wire be $L$.

At equilibrium, the tension in the wire is $T_0 = mg$.

The elongation of the wire at equilibrium is:

$\Delta L_0 = \dfrac{mgL}{AY}$

When the sphere passes through the lowest point (mean position) after being released from an angle $\theta$, its velocity $v$ is given by conservation of energy:

$v^2 = 2gL(1 - \cos\theta)$

The tension $T$ at the lowest point is:

$T = mg + \dfrac{mv^2}{L} = mg + 2mg(1 - \cos\theta)$

For the sphere to just avoid rubbing the floor, the maximum elongation of the wire at the lowest point must be $\Delta L = \Delta L_0 + h$, where $h$ is the equilibrium gap.

The tension can also be expressed using Hooke's law for this maximum elongation:

$T = \dfrac{YA\Delta L}{L} = \dfrac{YA}{L}\left(\dfrac{mgL}{AY} + h\right) = mg + \dfrac{YAh}{L}$

Equating the two expressions for tension $T$:

$mg + 2mg(1 - \cos\theta) = mg + \dfrac{YAh}{L}$

$2mg(1 - \cos\theta) = \dfrac{YAh}{L}$

$1 - \cos\theta = \dfrac{YAh}{2mgL}$

Given values are:
$Y = 2.0 \times 10^{11} \text{ N/m}^2$
$d = 1 \text{ mm} = 10^{-3} \text{ m} \Rightarrow A = \dfrac{\pi d^2}{4} = \dfrac{\pi}{4} \times 10^{-6} \text{ m}^2$
$h = 2 \text{ mm} = 2 \times 10^{-3} \text{ m}$
$m = 20 \text{ kg}$
$L = 4 \text{ m}$

Taking $g = 10 \text{ m/s}^2$:

$1 - \cos\theta = \dfrac{(2.0 \times 10^{11}) \times \left(\dfrac{\pi}{4} \times 10^{-6}\right) \times (2 \times 10^{-3})}{2 \times 20 \times 10 \times 4}$

$1 - \cos\theta = \dfrac{100\pi}{1600} = \dfrac{\pi}{16} \approx \dfrac{3.1416}{16} \approx 0.196$

$\cos\theta \approx 1 - 0.196 = 0.804$

$\theta = \cos^{-1}(0.804) \approx 36.4^{{}^{{}^{\circ}}}$