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 $$\frac{I_1}{I_2}$$ is $$\frac{x\pi^2}{3}$$, then the value of $$x$$ will be ______.

We need to find the ratio $$I_1/I_2$$ where $$I_1$$ is the moment of inertia of a rod about one end and $$I_2$$ is the moment of inertia of the same rod bent into a ring about its diameter.

For a uniform rod of mass M and length L about a perpendicular axis through one end: $$I_1 = \frac{ML^2}{3}$$

When the rod is bent into a ring, the circumference equals the length of the rod: $$2\pi r = L \implies r = \frac{L}{2\pi}$$

The moment of inertia of a ring of mass M and radius r about its diameter is $$I_2 = \frac{Mr^2}{2} = \frac{M}{2}\left(\frac{L}{2\pi}\right)^2 = \frac{ML^2}{8\pi^2}$$

Therefore, $$\frac{I_1}{I_2} = \frac{ML^2/3}{ML^2/(8\pi^2)} = \frac{8\pi^2}{3}$$ and comparing with $$\frac{x\pi^2}{3}$$ gives $$\frac{x\pi^2}{3} = \frac{8\pi^2}{3}$$ so $$x = 8$$.

The answer is 8.