Suspended from a nail in a wall, a closed circular wire experiences...

The time period of small oscillations for a physical pendulum is given by $T = 2\pi \sqrt{\dfrac{I}{mgd}}$, where $I$ is the moment of inertia about the axis of rotation and $d$ is the distance from the point of suspension to the center of mass.

For a circular wire (ring) of mass $m$ and radius $R$ suspended from a point on its circumference and oscillating in its own plane, the moment of inertia about the point of suspension is:

$I = I_{CM} + mR^2 = mR^2 + mR^2 = 2mR^2$

The distance from the point of suspension to the center of mass is $d = R$.

Substituting these values into the time period formula, we get:

$T = 2\pi \sqrt{\dfrac{2mR^2}{mgR}} = 2\pi \sqrt{\dfrac{2R}{g}}$

Given that $T = 2 \text{ s}$ and $g = \pi^2 \text{ m/s}^2$, we can substitute these into the equation:

$2 = 2\pi \sqrt{\dfrac{2R}{\pi^2}}$

$1 = \pi \dfrac{\sqrt{2R}}{\pi}$

$1 = \sqrt{2R}$

Squaring both sides:

$2R = 1$

$R = 0.5 \text{ m} = 50 \text{ cm}$