The moment of inertia of a uniform thin rod about a perpendicular...

Moment of inertia of rod rotating about an axis passing through one end is ${I}_{1} = \dfrac{m{l}^{2}}{3} ⋯$\left(1\right)$</p><p>Moment of inertia of ring rotating about one of its diameters is${I}_{2} = \dfrac{m{R}^{2}}{2}$.</p><p>Since the rod is bent into a ring, the circumference of the ring and the length of the rod are equal,$l = 2\piR$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/322b3509-32b7-46b3-8e3c-fa1057ddab24-image.png" style="width: 150px; height: 133px;" /></p><p>Then,${I}_{2} = \dfrac{m{R}^{2}}{2} = \dfrac{m{l}^{2}}{8{\pi}^{2}}$</p><p>Thus, the ratio$\dfrac{{I}_{1}}{{I}_{2}} = \dfrac{8{\pi}^{2}}{3}$, so$x = 8.