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

<p>To solve this problem, we need to understand the concept of moment of inertia. Moment of inertia is a measure of an object's resistance to rotational motion. It depends on the mass distribution of the object and the axis of rotation.</p> <p>For a continuous object like a semicircular ring, we can calculate the moment of inertia by integrating over the entire object. Here's how we can approach this problem:</p> <p>1. <strong>Divide the semicircular ring into small mass elements:</strong> Imagine the semicircular ring divided into infinitesimally small mass elements, each with mass $dm$.</p> <p>2. <strong>Calculate the moment of inertia of each element:</strong> The moment of inertia of each element about the axis passing through the center and perpendicular to the plane of the ring is given by $dI = dmR^2$, where R is the radius of the ring.</p> <p>3. <strong>Integrate to find the total moment of inertia:</strong> To find the total moment of inertia, we need to integrate $dI$ over the entire ring. This means integrating from $0$ to $\pi$ (the angle spanned by the semicircle) with respect to the angle $\theta$.</p> <p>4. <strong>Relate $dm$ to the total mass:</strong> Since the ring has a uniform mass distribution, we can express the mass of each element $dm$ as a fraction of the total mass $M$: $dm = \frac{M}{πR} Rd\theta = \frac{M}{\pi} d\theta$.</p> <p>Now, let's perform the integration:</p> <p>$$I = \int_{0}^{\pi} dI = \int_{0}^{\pi} dmR^2 = \int_{0}^{\pi} \frac{M}{\pi} d\theta R^2$$</p> <p>$$I = \frac{MR^2}{\pi} \int_{0}^{\pi} d\theta = \frac{MR^2}{\pi} [\theta]_{0}^{\pi}$$</p> <p>$I = \frac{MR^2}{\pi} [\pi - 0] = MR^2$</p> <p>Therefore, the moment of inertia of the semicircular ring about the given axis is $MR^2$. Comparing this to the given formula, we find that $x = \boxed{1}$.</p> <p></p>