JEE Main 2023 — Rotational Motion Question with Solution

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.

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:

1. Divide the semicircular ring into small mass elements: Imagine the semicircular ring divided into infinitesimally small mass elements, each with mass $$dm$$.

2. Calculate the moment of inertia of each element: 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 $$dI=dm{R}^2$$, where R is the radius of the ring.

3. Integrate to find the total moment of inertia: 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 $$\theta$$.

4. Relate $$dm$$ to the total mass: Since the ring has a uniform mass distribution, we can express the mass of each element $$dm$$ as a fraction of the total mass $$M$$: $$dm=\frac{M}{\pi R}Rd\theta =\frac{M}{\pi }d\theta$$.

Now, let's perform the integration:

$$I={\int }_0^{\pi }dI={\int }_0^{\pi }dm{R}^2={\int }_0^{\pi }\frac{M}{\pi }d\theta R^2$$

$$I=\frac{M{R}^2}{\pi }{\int }_0^{\pi }d\theta =\frac{M{R}^2}{\pi }[\theta {]}_0^{\pi }$$

$$I=\frac{M{R}^2}{\pi }[\pi -0]=M{R}^2$$

Therefore, the moment of inertia of the semicircular ring about the given axis is $$M{R}^2$$. Comparing this to the given formula, we find that $$x=\boxed{1}$$.