JEE Physics 2023 Question with Solution

Given: The moment of inertia of the disc about any diameter is $\frac{MR^2}{4}$.

Find: The value of $x$ if the moment of inertia about an axis normal to the disc and passing through a point on its edge is $\frac{x}{2}MR^2$.

Use the perpendicular axis theorem first.

So, the moment of inertia about the axis through the center and perpendicular to the disc is $\frac{MR^2}{2}$.

Now apply the parallel axis theorem for an axis normal to the disc and passing through its edge:

Substitute $I_c = \frac{MR^2}{2}$:

Given that

Comparing,

Hence,

Therefore, the value of $x$ is $3$.

Given: $I_d = \frac{MR^2}{4}$ for a diameter of the disc.

Find: The numerical value of $x$ in $\frac{x}{2}MR^2$.

For a circular disc, two mutually perpendicular diameters lie in the plane of the disc. By the perpendicular axis theorem, the moment of inertia about the axis through the center and perpendicular to the plane is the sum of the moments of inertia about these two diameters.

The required axis is parallel to this central normal axis and passes through the edge of the disc. The separation between these two axes is $R$. By the parallel axis theorem,

Now compare this with the given form:

So,

Therefore, the correct numerical answer is $3$.