The momentum of inertia of a uniform thin rod about a perpendicular axis passing through on end is I₁. The same rod is into a ring and its moment of inertia about a diameter is I₂. If $\frac{I_1}{I_2}is\frac{X{\pi }^2}{3},$ then the value of x will be__________.

According to question, we can write

$\mathcal{l}=2\pi r\Rightarrow r=\frac{\mathcal{l}}{2\pi }$

$I_1=\frac{m{\mathcal{l}}^2}{3},and{I}_2=\frac{m{r}^2}{2}=\frac{m{\mathcal{l}}^2}{8{\pi }^2}$

$\Rightarrow \frac{I_1}{I_2}=\frac{m{\mathcal{l}}^2}{3}\times \frac{8{\pi }^2}{m{\mathcal{l}}^2}=\frac{8{\pi }^2}{3}$

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