JEE Physics 2025 Question with Solution

Given: A circular ring has mass $M$ and diameter $r$.

Find: The moment of inertia about a tangential axis lying in the plane of the ring.

For a ring, if the diameter is $r$, then the radius is

$$R = \frac{r}{2}$$

The moment of inertia of a ring about a diametral axis through the center and lying in its plane is

$$I_{\text{center, diametral}} = \frac{1}{2}MR^2$$

Using the parallel axis theorem for the tangential axis at distance $R$ from the center,

$$I_{\text{tangent, in-plane}} = I_{\text{center, diametral}} + MR^2$$ $$I_{\text{tangent, in-plane}} = \frac{1}{2}MR^2 + MR^2 = \frac{3}{2}MR^2$$

Now substitute $R = \frac{r}{2}$:

$$I_{\text{tangent, in-plane}} = \frac{3}{2}M\left(\frac{r}{2}\right)^2 = \frac{3}{8}Mr^2$$

Therefore, the moment of inertia is $\frac{3}{8}Mr^2$. The correct option is A.

The solution states the correct option is B, but the listed listed options begin at (2), so the worked value $\frac{3}{8}Mr^2$ corresponds to the first available option in the recorded data.

Given: Mass of ring $= M$, diameter $= r$.

Find: Moment of inertia about a tangent in the plane of the ring.

Concept used: Parallel Axis Theorem.

For a circular ring of radius $R$,

$$I_{\text{center, diametral}} = \frac{1}{2}MR^2$$

The tangential axis in the plane is parallel to this diametral axis and is displaced by distance $R$. So,

$$I = I_{\text{center}} + Md^2$$

with

$$d = R$$

Hence,

$$I = \frac{1}{2}MR^2 + MR^2$$ $$I = \frac{3}{2}MR^2$$

Since the given diameter is $r$,

$$R = \frac{r}{2}$$

Substituting,

$$I = \frac{3}{2}M\left(\frac{r}{2}\right)^2$$ $$I = \frac{3}{8}Mr^2$$

Thus, the required moment of inertia is $\frac{3}{8}Mr^2$.