Question 45 of 280

The moment of inertia of a uniform thin rod about a perpendicular axis passing through one end is ${I}_{1}$. The same rod is bent into a ring and its moment of inertia about a diameter is ${I}_{2}$. If $\dfrac{{I}_{1}}{{I}_{2}}$ is $\dfrac{x{\pi }^{2}}{3}$, then the value of $x$ will be ______.

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Answer: 8

Moment of inertia of rod rotating about an axis passing through one end is ${I}_{1}=\dfrac{m{l}^{2}}{3}\cdots \left(1\right)$
Moment of inertia of ring rotating about one of its diameters is ${I}_{2}=\dfrac{m{R}^{2}}{2}$.
Since the rod is bent into a ring, the circumference of the ring and the length of the rod are equal, $l=2\pi R$

Then, ${I}_{2}=\dfrac{m{R}^{2}}{2}=\dfrac{m{l}^{2}}{8{\pi }^{2}}$
Thus, the ratio $\dfrac{{I}_{1}}{{I}_{2}}=\dfrac{8{\pi }^{2}}{3}$, so $x=8$.