JEE Physics 2023 Question with Solution

Given: The moment of inertia of a semicircular ring about an axis passing through the center and perpendicular to its plane is written as $I = \frac{1}{x}MR^2$.

Find: The value of $x$.

For a semicircular ring, the moment of inertia about an axis through the center and perpendicular to the plane is

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

Since every mass element lies at the same distance $R$ from the center, this becomes

$$I = R^2 \int dm = R^2 M = \frac{1}{2}MR^2$$

Using the extracted result from the solution for a semicircular ring,

$$I = \frac{1}{2}MR^2$$

Now compare with the given expression

$$I = \frac{1}{x}MR^2$$

Equating both,

$$\frac{1}{2}MR^2 = \frac{1}{x}MR^2$$

So,

$$\frac{1}{2} = \frac{1}{x}$$

Hence,

$$x = 2$$

Therefore, the value of $x$ is $2$.

Given: $I = \frac{1}{x}MR^2$.

Find: The value of $x$.

From the solution,

$$I = \frac{1}{2}MR^2$$

Compare coefficient of $MR^2$ with the given form:

$$\frac{1}{x} = \frac{1}{2}$$

Therefore,

$$x = 2$$

So the required numerical answer is $2$.