The moment of inertia of a semicircular ring about an axis, passing through the center and perpendicular to the plane of ring, is $$\dfrac{1}{x}$$ MR$$^2$$, where R is the radius and M is the mass of the semicircular ring. The value of x will be ______.

A semicircular ring has all its mass distributed at a constant distance $$R$$ from the center.

The moment of inertia about an axis passing through the center and perpendicular to the plane of the ring is:

$$ I = \int r^2\,dm $$

Since every element of mass is at the same distance $$R$$ from the center:

$$ I = R^2 \int dm = MR^2 $$

Comparing with $$\frac{1}{x}MR^2$$: $$\frac{1}{x} = 1 \implies x = 1$$.

The value of $$x$$ is 1.