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(D) For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$.Here,$I$ is the moment of inertia about the pivot point,$m$ is

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$2$

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(D) For a physical pendulum,the time period is given by $T = 2\pi \sqrt{\frac{I}{mgd}}$.Here,$I$ is the moment of inertia about the pivot point,$m$ is the mass,and $d$ is the distance from the center of mass to the pivot.For a ring pivoted at its circumference,the moment of inertia about the pivot is $I = I_{cm} + md^2 = mR^2 + mR^2 = 2mR^2$.The distance $d$ from the center to the pivot is $R$.Substituting these into the formula: $T = 2\pi \sqrt{\frac{2mR^2}{mgR}} = 2\pi \sqrt{\frac{2R}{g}}$.Given the diameter $D = 2R = 1 \ m$,so $R = 0.5 \ m$.Taking $g = 9.8 \ m/s^2$ or approximately $\pi^2 \approx 9.8$ for simplification,$T = 2\pi \sqrt{\frac{1}{g}}$.If we use $g = \pi^2$,then $T = 2\pi \sqrt{\frac{1}{\pi^2}} = 2 \ s$.

$A$ ring whose diameter is $1 \ m$ oscillates simple harmonically in a vertical plane about a nail fixed at its circumference. The time period will be ....... $s$.

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