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

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$\sqrt{\frac{3}{2 \pi}} T$

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(D) For a physical pendulum,the time period is given by $T = 2 \pi \sqrt{\frac{I}{mgd}}$,where $I$ is the moment of inertia about the pivot,$m$ is the mass,$g$ is acceleration due to gravity,and $d$ is the distance from the pivot to the center of mass $(COM)$.For the rod of length $\ell$ suspended from one end:$I = \frac{m\ell^2}{3}$ and $d = \frac{\ell}{2}$.$T = 2 \pi \sqrt{\frac{m\ell^2/3}{mg\ell/2}} = 2 \pi \sqrt{\frac{2\ell}{3g}} \quad \dots(i)$When the wire is bent into a circle of radius $R$,the circumference is $\ell = 2 \pi R$,so $R = \frac{\ell}{2 \pi}$.For a ring suspended from a point on its circumference,the moment of inertia about the pivot is $I' = I_{cm} + mR^2 = mR^2 + mR^2 = 2mR^2$ (using parallel axis theorem).The distance from the pivot to the $COM$ is $d' = R$.$T' = 2 \pi \sqrt{\frac{2mR^2}{mgR}} = 2 \pi \sqrt{\frac{2R}{g}} \quad \dots(ii)$Substituting $R = \frac{\ell}{2 \pi}$ into $(ii)$:$T' = 2 \pi \sqrt{\frac{2}{g} \cdot \frac{\ell}{2 \pi}} = 2 \pi \sqrt{\frac{\ell}{\pi g}}$.Dividing $(ii)$ by $(i)$:$\frac{T'}{T} = \frac{2 \pi \sqrt{2R/g}}{2 \pi \sqrt{2\ell/3g}} = \sqrt{\frac{2R}{g} \cdot \frac{3g}{2\ell}} = \sqrt{\frac{3R}{\ell}} = \sqrt{\frac{3(\ell/2\pi)}{\ell}} = \sqrt{\frac{3}{2 \pi}}$.Thus,$T' = \sqrt{\frac{3}{2 \pi}} T$.

$A$ long stiff uniform wire of length $\ell$ is suspended from one end. The time period of oscillations of the wire is $T$. If the wire is now bent into a circle and suspended from a knife edge so that it can oscillate freely in the plane of the ring,its time period will be

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