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**Hint**We will use the general formula for the moment of inertia of a circular ring, which is equal to $m{r^2}$, and the moment of inertia of a rod, which is equal to $\dfrac{{m{l^2}}}{{12}}$. We will relate the length of the rod with the circumference of the ring since both are the same physical length.

**Complete step by step answer**

The formula for the moment of inertia of the ring about its own axis is $m{r^2}$, where $r$ is the radius and $m$ is the mass of the ring.

Therefore we have ${I_1} = m{r^2}$.

When this same ring is cut into a rod, the length of the rod becomes equal to the perimeter of the ring.

This means that

$2\pi r = l$,

where $l$ is the length of the rod.

Thus the new moment of inertia, the moment of inertia of the rod about an axis perpendicular to the length of the rod and passing through its center is given by the general formula

$\dfrac{{m{l^2}}}{{12}}$,

where $m$ is the mass of the rod and

$l$ is the length of the rod.

Here we will substitute the value of the length of the rod with $l = 2\pi r$.

Therefore the moment of inertia becomes

${I_2} = \dfrac{{m{{(2\pi r)}^2}}}{{12}} = \dfrac{{m{{(\pi r)}^2}}}{3}$.

We need to calculate the ratio between the moment of inertias of the ring and the rod, i.e. $\dfrac{{{I_1}}}{{{I_2}}} = \dfrac{{m{r^2}}}{{\dfrac{{m{{(\pi r)}^2}}}{3}}}$.

$ \Rightarrow \dfrac{{{I_1}}}{{{I_2}}} = \dfrac{3}{{{\pi ^2}}}$

This is the required ratio of the moments of inertias, i.e. the answer is $\dfrac{3}{{{\pi ^2}}}$.

**Hence option the correct answer is option (B).**

**Note**Here we have considered the mass moment of inertia and not the area moment of inertia. The mass moment of inertia is easier and required to be calculated here since nothing is given about the area and we will consider the thin rod and ring to be effectively a thin straight line.

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